math perm and comb

Mathematics / P & C, Binomial Theorem www.locuseducation.org

01. Fundamental Principle of Counting 01 - 04

02. Introduction To Permutations 05 - 07

03. Introduction To Combinations 08 - 09

04. Applications of Basics 10 - 26

05. More Applications 27 - 59

06. Appendix : Binomial Theorem 60 - 86

CONCEPT NOCONCEPT NOCONCEPT NOCONCEPT NOCONCEPT NOTESTESTESTESTES

P & C,

Binomial Theorem


LOCUSLOCUSLOCUSLOCUSLOCUS 1

P & C,Binomial Theorem

This chapter is one of the most interesting chapters that we’ll study at this level. The beauty and challenge ofthis branch of mathematics lies in the innumerous tricks and mathematical artifices that abound in this subject.Clarity of thought, more than any thing else, is what is required to understand the subject properly. Also, you’ddo best if you refrain from memorising any formulae or particular cases here; concentrate on building a logicalapproach, solving everything from first principles.

The main objective of this chapter is to count. Given a set U of things or objects or persons ( or whatever), weneed to arrange a subset S of U (according to some constraints) or select a subset S of U (again, according tocertain criteria). In fact, we are actually interested in counting the number of such arrangements or selections.Read the following questions:

“From a team of 15 cricket players, how can we select a playing team of 11players?

“There are 20 people whom we need to seat in 2 rows of 10 seats each. How many ways exist of doing so?

“From a deck of 52 playing cards, in how many ways can we select two red cards and three black cards?”

“How many rectangles exist on a standard 8×8 chessboard?”

“How many factors does 144000 have? In general, how many factors does a natural number N have?”

These are some of the many types of questions which we’ll learn to solve in this chapter.

We’ll build a systematic approach to deal with such counting issues. To really appreciate the beauty of thesolving techniques that we’ll develop, you are urged to try out each and every question that we solve here, onyour own first, and only then look at the solution. Only this approach will help you solve counting questionselegantly.

Section - 1 Section - 1FUNDAMENTAL PRINCIPLE OF COUNTING

The fundamental principle of counting is so fundamental that you already must have used it practically a lotmany times without realising it. In other words, this principle is already programmed into our minds. A logical,step-by-step application of this principle gives rise to the entire subject of permutations and combinations.

Suppose you have 4 boys and 3 girls. From this group of 7, you want to select a couple ( a boy and a girl).How many ways exist of forming this couple?

Let us label the boys as 1 2 3 4, , and B B B B and the girls as 1 2 3, and G G G .



There are 4 ways to choose a boy. Once you’ve chosen a boy, say 2B , there are now 3 ways to choose a girl.In other words, the boy 2B can form the couple 2 1 2 2 2 3( ), ( ) or ( )B G B G B G . Similarly, for every otherboy, there exist 3 girls with whom that boy can be paired. Thus, the total number of ways in which a pair canbe formed is:

( ) ( )No. of ways of selecting a boy No. of ways of selecting a girlN = ×

4 3= ×

12=

The most crucial aspect in this calculation is that you must realise that the task of selecting a boy is independentof the task of selecting a girl. This means that which boy you select has no effect what so ever on which girl youselect; the selection of a boy and that of a girl are independent of each other.Another subtle point must be made here. There are 4 ways to select a boy. These 4 ways are mutuallyexclusive. This means that a selection of any particular boy, once made, rules out the selection of the other 3boys. Similarly, there are 3 mutually exclusive ways to select a girl.Let us consider another example now. We need to travel from New Delhi, India to Fiji Islands (in the SouthPacific Ocean). We must change flights, first at Singapore and then at Sydney, Australia. There are 6 differentflights available from New Delhi to Singapore, 5 from Singapore to Sydney and 3 from Sydney to Fiji.

New Delhi Singapore Sydney Fiji Islands

3 flights5 flights6 flights

Fig - 1

How may ways exist of making a flight plan from New Delhi to Fiji?Assume that we are in Singapore. From Singapore, we have 5 available flights to Sydney (5 mutually exclusiveways). For each of these 5 flights, we have 3 further ways (flights) from Sydney to Fiji. Thus, we have a totalof 5 × 3 = 15 ways of travelling from Singapore to Fiji.Now, how many ways do we have to reach Singapore from New Delhi in the first place? 6 flights. For each ofthese 6 flights, we have 15 further ways of reaching Fiji From Singapore (as we just calculated in the precedingparagraph). Thus, the total number of ways N of travelling from New Delhi to Fiji is 6 × 15 = 90. In otherwords,

( ) ( ) ( )No. of flights from New No. of flights from No. of flights from Delhi to Singapore Singapore to Sydney Sydney to FijiN = × ×

Intuitively easy, isn’t it?Let us now generalise the results of these two examples into our fundamental principle of counting.Consider the set of tasks 1 2 3{ , , ... }nT T T T which are all independent of each other. This means that task iThas “no relation ” to task if ;jT i j≠ the choice of how to accomplish task iT has thus “no effect” on thechoice of how to accomplish talk jT for i j≠ , (you’ll realise the meaning of “no relation” and “no effect” morespecifically later). Task iT can be accomplished in ik mutually exclusive ways.



The fundamental principle of counting says that the set of tasks 1 2 3{ , , ... }nT T T T can be accomplishedin 1 2 3 ... nk k k k× × × × ways.Here are more examples that illustrate this simple yet extremely powerful principle. A proper understanding ofthis principle is absolutely essential for the subject of counting to be fully comprehensible.

• Think of a standard six-faced die, with the markings 1, 2, 3, 4, 5 and 6. Suppose that you havetwo such dice. When these two dice are thrown call the number that shows up on the first die xand the one on the second die y. How many pairs ( , )x y are possible?There are 6 mutually exclusive ways in which a number can show up on the first die. Similarly, 6such ways exist for the second die. The fundamental principle of counting says that the totalnumber of possible pairs is 6 × 6 = 36.

• Suppose that we have a (unlimited) supply of the letters , , , , ,A B C D E F available with us.How many 5-letter chains can we form using these letters?Imagine 5 blank spaces for the 5-letter chain that we are supposed to form (numbered 1 to 5):

1 2 3 4 5Fig - 2

• We now have 5 tasks at hand; each task correspond to filling a space in Fig. 1. Realise that these5 tasks are independent of each other. For example, what you fill in place 2 has no effect on whatyou fill in place 5 since you’ ve been assured an unlimited supply of letters. Each task can beaccomplished in 6 possible ways. For example, you can fill place -1 in 6 possible ways : with

, , , , , orA B C D E F . Thus, by the fundamental principle of counting, the total number of waysto form the 5-letter chain would be 56 6 6 6 6 6× × × × = .

• Now consider the scenario when you don’t have an unlimited supply of letters available. Supposethat you have only one of each of the 5 letters available. How many 5-letter chains can be formednow?You must realise that this limited-supply-of-letters situation is very different from the previous one.In this situation, once you fill a particular place with a particular letter, you are left with one letterless to choose from, from the remaining places.Let us start with filling the places from left to right. To fill place -1, w e have 6 letters to choosefrom, but once we’ve filled place-1 , we now have only 5 letters to choose from, to fill place-2.Continuing in this way, place-3, place - 4 and place -5 can then be filled in 4, 3 and 2 waysrespectively.

6 ways to fill 5 ways to fill 4 ways to fill 3 ways to fill 2 ways to fill this place this place this place this place this place

1 2 3 4 5

! ! ! ! !

Fig - 3

The fundamental principle of counting tells us that the total number of chains in this case will be6 5 4 3 2 720× × × × = . You might think that in this case filling up place-i and place-j are notindependent tasks. In other words, how we fill place-i has a certain relation to how we



fill place-j . For example, if we fill place-1 with A, we cannot fill any other place with A and thus,how we fill place -1 has a definite effect on how we are able to fill the other places. This isdefinitely true. But when we talk about the independence of two events, it is in a different sense:Let event X be filling place-1 and event Y be filling place-2 after event X has been accomplished,i.e. after place-1 has been filled. There are 6 ways of accomplishing X and 5 ways of accomplishingY. Which 5 letters contribute to event Y (i.e which of the 5 remaining letters can we choose forplace -2) definitely depends on how event X was accomplished, but the event Y is itself the choiceof one of the 5 symbols contributing to event Y. The number of ways in which this choice can bemade is still independent of how event X was accomplished. Whatever selection was made forevent X, the number of ways in which Y can be accomplished still remains 5. This is the sense thatshould be attached to the phrase “independent events”.

____________________________________________________________________________________

These three examples should have given you a pretty good idea about three concepts : mutually exclusiveevents, independent events and the fundamental principle of counting. Ponder over these examples for a bitlonger and think up examples of you own till your feel very comfortable with these new concepts.



Section - 2 Section - 1INTRODUCTION TO PERMUTATIONS

We now formally study the concept of permutations, by generalising the last example in the preceding section.

The fundamental issue in Permutations is the arrangement of things. In the last example, we had 6 letters and5 places where we could arrange (5 of) those 6 letters, We calculated that there are 720 ways of arrangingthose 6 letters, taken 5 at a time. In mathematical terminology, we calculated as 720 the number of permutations(arrangements) of 6 letters taken 5 at a time.

Let us generalise this: Suppose we have n people. It we had n seats available to seat these people, the totalnumber of ways to do so would be (by the logic discussed in the preceding section) ( 1) ( 2) ... 1n n n× − × − × × .This quantity is denoted by n!.

! ( 1) ( 2) ... 1n n n n= × − × − × ×

Suppose now that we have only r seats, where r < n. The total number of ways now wouldbe ( 1) ( 2) ... ( 1)n n n n r× − × − × × − + . This is the number of ways of permuting n things, taken r at a time,and the notation used for this number is n

rP . Thus :

( 1) ( 2) ... ( 1)nr n n rP n n= × − × − × × − +

( 1) ( 2) ... ( 1) ( ) ...1( ) ... 1

n n n n r n rn r

× − × − × × − + × − ×=− × ×

!( )!

nn r−

=

Thus, using the fundamental principle of counting, you see that we’ve been able to calculate the value of nrP .

Finally, note that !nnP n=

Lets apply this discussion to an example:

Example – 1

Find the number of permutations of the word EDUCATION which

(a) consist of all the letters of this word (b) start with E and end with N

(c) contain the string “CAT” (d) have the letters C, A and T occurring together

(e) start and end with vowels (f) have no two vowels occurring together.

Solution: (a) We have 9 letters and we want to permute all of them. The required number of arrangementswould be 9

9 9!P =



(b) Fix E at the start and N at the end of the word

E N

Fig - 4

We now have 7 places which we need to fill using the remaining 7 letters. The number of suchpermutations will be 7

7 7!P = .

(c) We want all those permutations in which the string “CAT” occurs. Let us treat “CAT” as asingle letter/object since this is what we want - we want “CAT” to appear as a single entity.We now have the following objects (letters) which we need to permute:

E D U “CAT ” I O N

These are in total 7 objects. Thus, they can be permuted in 77 7!P = ways. This is the number

of permutations that contain the string “CAT ”.

(d) We now need the permutations which contain the letters C, A and T occurring together. Thismeans that they can occur together in any order. There are 3! = 6 ways in which C, A and Tcan be arranged among themselves, namely CAT, CTA, ACT, ATC, TAC, and TCA.

Now consider from the previous part all those permutations in which the string “CAT” occurs.These are 7! in number. Corresponding to each such permutation, we can have 6 permutationsin which the constraint is that C, A and T occur together. This is because the string “CAT” canitself be permuted in 6 ways as described above. For example, consider the permutation“EDUCATION”; to this permutation will correspond 6 permutations in which C, A and Toccur together:

EDU CAT IONEDU CTA IONEDU ACT IONEDU CAT IONEDU ATC IONEDU TAC IONEDU TCA ION

→

The required number of permutations is therefore 7!×3!.

(e) In the word EDUCATION, there are a total of 5 vowels. We want to have the starting andending letters of our permutations as vowels. Let us first arrange the starting and ending lettersand then fill in the rest of the 7 places. Out of the 5 vowels available to us, the first and the lastplace can be filled in 5

2P places. Once we’ve fixed the first and the last letters, the remaining7 places can be filled in 7

7 7!P = ways. Thus, the total number of required permutations is5

2 7!P × .



(f) There are 5 vowels and 4 consonants in the word EDUCATION. We require permutations inwhich no two vowels occur together. We can ensure such permutations by first fixing the 4consonants and then arranging the 5 vowels in the 5 possible places that arise as depicted inthe following figure.

D C T N

Fig - 5

Convince yourself that any arrangement of the 5 vowels in the 5 blank spaces above willcorrespond to a permutation with no two vowels together.

The number of ways of arranging the 4 consonants is 44 4!P = . After the consonants have

been arranged, the number of ways of arranging the 5 vowels in the 5 blank spaces as depicted

in Fig.5 is 5!. Thus, the required number of permutations is 4! × 5!

____________________________________________________________________________________

It would be a good exercise for you to construct more of such examples on your own and solve them. You cantake an arbitrary word and find the number of permutations of that word, that satisfy any particular conditionthat you can think of.



Section - 3 Section - 1INTRODUCTION TO COMBINATIONS

In the previous section, we encountered permutations, which correspond to arrangement ofobjects /things/ entities. In this section, we will encounter combinations which correspond to selection of things(and not their arrangement). We do no intend to arrange things. We intend to select them. For example,suppose we have a team of 15 cricket players. We intend to select a playing team of 11 out of these 15players. Thus, we want the number of ways in which we can select 11 players out of 15 players. (We are notinterested in arranging those 11 players in a row - only the group/ combination of those 11 players matters).

Let us make this concept more specific. Suppose we have a set of 6 letters { }, , , , ,A B C D E F . In howmany ways can we select a group of 3 letters from this set? Suppose we had to find the number of arrangementsof 3 letters possible from those 6 letters. That number would be 6

3P . Consider the permutations that containthe letters A, B and C. These are 3! = 6 in number, namely ABC, ACB, BAC, BCA, CAB and CBA.

Now, what we want is the number of combinations and not the number of arrangements. In other words, the6 permutations listed above would correspond to a single combination. Differently put, the order of things isnot important; only the group/combination matters. This means that the total number of combinations of 3letters from the set of 6 letters available to us would be 6

3 / 3!P since each combination is counted 3! times inthe list of permutations. Thus, if we denote the number of combinations of 6 things taken 3 at a time by 6

3C , wehave

66 3

3 3!PC =

In general, suppose we have n things available to us, and we want to find the number of ways in which we canselect r things out of these n things.

We first find the number of all the permutations of these n things taken r at a time. That number would be nrP .

Now, in this list of nrP permutations, each combination will be counted r! times since r things can be permuted

amongst themselves in r! ways. Thus, the total number of combinations of these n things, taken r at a time,denoted by n

rC , will be

!! !( )!

nn r

rP nCr r n r

= =−

You should now be able to appreciate the utility of the fundamental principle of counting. Using only a step-by-step application of this principle, we have been able to obtain an expression for .n

rC As we progress throughthe chapter, you’ll slowly realise that each and every concept that we discuss and each and every expressionthat we obtain follows logically as a consequence of this simple principle.



Example – 2

Consider the word EDUCATION.

(a) In how many ways can we select a pair of letters consisting of a vowel and a consonant ? Bothvowels? Both consonants?

(b) In how many ways can a 5-letter selection be made with more than 2 vowels?

Solution: (a) We have 5 vowels and 4 consonants available in the word EDUCATION. A vowel can beselected in 5 ways while a consonant can be selected in 4 ways. Thus, a pair consisting of avowel and a consonant can be selected in 5 × 4 = 20 ways. Two vowels can be selected in5

2C ways while two consonants can be selected in 42C ways.

(b) Since we want more than 2 vowels, we could have 3, 4 or 5 vowels in our 5-letter selection.

Suppose we select 3 vowels for our 5-letter group. This can be done in 53C ways. The two

remaining letters (consonants) can be selected in 42C ways. Thus, 5-letter groups containing

3 vowels can be formed in 5 43 2C C× ways. Similarly, if we had 4 vowels, the possible number

of groups would be 5 44 1C C× . There’s only 1 group possible containing (all) the 5 vowels.

Thus, the number of required selections is

5 4 5 43 2 4 1 1C C C C× + × +

____________________________________________________________________________________

Now that we’re done with the introductions, lets move on and see some really interesting and diverse applicationsof the basic concepts covered till now.



Section - 4 Section - 1APPLICATIONS OF BASICS

Our approach in the applications of the basic concepts of counting will be to keep everything really basic - andtry to do everything from a first principles approach. Memorization of formulae will do no good! A good dealof thinking is required.

Lets start with proving certain combinatorial assertions.

Example – 3

Prove that n nr n rC C −=

Solution: We can easily prove this relation using the expression we obtained for nrC :

! !!( )! ( )!( ( ))!

n nr n r

n nC Cr n r n r n n r −= = =

− − − −

However, what we would really like is not an analytical justification like the one above but alogical justification, that involves no mathematical manipulations and is instead purely based on theinterpretation of n

rC .For this example, let us discuss such a logical justification in a good amount of detail.Suppose you have a group of 6 letters, say {A, B, C, D, E, F}. Out of this group, you want tocount the number of subsets containing 4 letters. Consider a particular selection of 4 letters, say{B, D, E, F}. This selection can equivalently be obtained if we say that we exclude the group{A, C} from our original group of 6 letters. This means that each selected group of 4 letterscorresponds to an excluded group of 2 letters. The number of (selected) 4-letter groups willtherefore equal the number of (excluded) 2-letter groups, or in other words, to count the numberof 4 letter groups, we can equivalently count the number of 2-letter groups. Thus,

6 64 2C C=

Generalising this logic, we obtain n nr n rC C −=

Example – 4

Prove that 1 11

n n nr r rC C C− −

−= +

Solution: You can prove this assertion yourself very easily analytically. Let us discuss a logical justification.

The left hand side of this assertion says that we have a group of n people out of which we want toselect a subset of r people; more precisely, we want to count the number of such r-subsets.

Fix a particular person in this group of n people, say person X. Now, all the r-groups that we formwill either contain X or not contain X. These are the only two options possible.



To count the number of groups that contain X, we proceed as follows: we already have X; weneed (r – 1) more people from amongst (n – 1) people still available for selection. Thus, suchgroups will be 1

1n

rC−− in number.

For the number of groups that do not contain X we need to select r people from amongst (n –1)options available. Therefore, such groups are 1n

rC− in number. The total number of r-groups are

hence 1 11

n nr rC C− −− + in number, which is the same as n

rC . With this example, you should begin

to realise the beauty of the logic and skill required in this subject.

Example – 5

Prove that 1 11

n n nr r rP P r P− −

−= +

Solution: The analytical justification is again very straightforward and is left to you as an exercise.

The left hand side of this assertion says that we need to count the number of arrangements of npeople, taken r at a time.

We again fix a particular person, say person X. All the possible r-arrangements will either containX or not contain X. These are the only two (mutually exclusive) cases possible.

If we do not keep X in our permutation, we have r people to select from a potential group of(n – 1) people. The number of arrangements not containing X will therefore be 1 .n

rP−

To count the number of permutations containing X, we first seat X in one of the r seats available.This can be done in r ways. The remaining (r – 1) seats can be filled by (n –1) people in 1

1n

rP−−

ways. Thus, the number of arrangements containing X is 11

nrr P−−⋅ .

These arguments prove that 1 11

n n nr r rP P r P− −

−= + ⋅

Example – 6

(a) Prove that 11

n nr r

nC Cr

−−= (b) Prove that 1

1n nr r

n rC Cr −

− +=

Solution: (a) Let us consider this assertion in a particular example, say with n = 6 and r = 4. This will makethings easier to understand.

Our purpose is to select 4 people out of 6 people, say the set {A, B, C, D, E, F}. To select agroup of 4, we can first select a single person : this can be done in 6 ways. The rest of the 3people can now be selected in 5

3C ways. The total number of groups possible would thus be5

36 C× . But some careful thought will show that we have ‘overcounted’, doing the calculationthis way.



Suppose that in our first step, we select A. While selecting the remaining 3 persons, we thenselect B, C, D thus forming the group {A, B, C, D}. But this same group would have beenformed had we selected B in our first step and A, C, D in the second step, or C in the first stepand A, B, D in the second step, or D in the first step and A, B, C in the second step. We havethus counted the group {A, B, C, D} 4 times in the figure 5

36 .C× The actual number of

groups will hence be 5

36 .4

C×

We now generalise this: to select r people out of a group of n, we first select one person; thiscan be done in n ways. The remaining (r – 1) persons can be selected in 1

1n

rC−− ways. The

total number of r-groups thus becomes –11

nrn C −× . However, as described earlier, in this

figure each group has been counted r times. The actual number of r-groups is therefore

11

nn r

rn CC

r

−−×=

(b) The logic for this part is similar to that of part - (a)

To select r people out of n, we first select (r – 1) people out of n. This can be done in 1n

rC −ways. We now have ( 1) ( 1)n r n r− − = − + persons remaining for selection out of which wehave to choose 1 more person. This can therefore be done in (n – r + 1)ways. The totalnumber of r-groups thus becomes 1( 1) .n

rn r C −− + ×

However, each r-group has again been counted r times in this figure (convince yourself aboutthis by thinking of a particular example). The actual number of r-group is thus

1( 1) nn r

rn r CC

r−− + ×=

Example – 7Example – 7

Prove that 0 1 2 .......... 2n n n n nnC C C C+ + + + =

Solution: This looks like a tough one! Lets first interpret what the left side means. 0nC is the number of

ways in which we can select ‘nothing’ out of n things (there will obviously be only one such way:that we do nothing!). 1

nC is the number of ways in which we can select 1 thing out of n. nrC is the

number of ways of selecting r things out of n. We want the value of the sum 0

nn

rr

C=∑ , which is the

number of all groups possible of any size what so ever. Thus, our selection could be any size from0 to n (both inclusive); what we want is the total number of selections possible.

For example, consider the set {A, B, C}. The set of all possible selections that can be made from this setis { } { } { } { } { } { } { }{ }, , , , , , , .A B C AB AC BC ABCφ Thus, 8 total different selections are possible

(note that 38 2= )



To count the total number of selections possible if we have n persons, we adopt an individual’sperspective. An individual can either be or not be in our selection. Thus, we have two choices withrespect to any individual; we either put him in our group or do not put him in our group.

These two choices apply to every individual. Also, choosing or not choosing any individual isindependent of choosing or not choosing another. Thus, the total number of ways in which an

arbitrary number of individuals can be selected from n people is times

2 2 2 ........ 2 2n

n× × × × ="###$###%

This proves that

02

nn n

ii

C=

=∑

Example – 8

Prove that 0 1 1 2 2 0.........n m n m n m n n n mr r r r rC C C C C C C C C+

− −= + + + +

Solution: We interpret the left hand side as the number of ways of select r people out of a group of (n + m)people.

Let this group of (n + m) people consist of n boys and m girls. A group of r people can be madein the following ways:

0

1 1

2 2

1 1

0

1 0 boys, girls2 1 boy, ( 1) girls3 2 boys, ( 2) girls.... ( 1) boys, 1 girl

( 1) boys, 0 girsl

n mr

n mr

n mr

n mr

n mr

C CrC Cr

r C C

r r C Cr r C C

−

−

−

×

×−− ×

− ×+ ×

Sl. Group contains No. of ways possible

This table is self-explanatory. The ( r + 1) types of groups that have been listed are mutuallyexclusive. Thus, the total number of r-groups is

0 1 1 0........m m mn m n n nr r r rC C C C C C C+

−= + + +



Example – 9

(a) Consider the set of letters { }, , , , , .a a a b c d How many permutations of this set exist?

(b) Consider the set of letters { }, , , , , , , .a a a b b c d e How many permutations of this set exist?

(c) Generalise the results of the previous two parts. If we have a set of ( ).....m n p+ + + things wherea particular thing, say X, is repeated m times, Y is repeated n times, Z is repeated p times, and soon, find the number of permutations of this set.

Solution: The main issue now is that we have a repetition of things. If the things we have were all different,the number of permutations would have been easy to calculate. But now, with repetition of things,the number of permutations will change (it will actually decrease, if you think about it carefully).Let us calculate the permutations in this case from first principles(a) We have 3 (repeated) “a” letters, a “b”, a “c” and a “d”. In all, 6 letters. Consider, for a

moment, that the 3 “a” s are all different. Let us denote the 3 different “a”s by a1, a2 and a3.Our set of letters is now { }1 2 3, , , , , .a a a b c d The number of permutations of this set is simply6

6 6!P =If we list down all these 6! permutations, we will see that 6 permutations from this list willcorrespond to only one permutation, had the “a”s been all the same. Why?

Consider any particular permutation with the “a” s all different, say { }1 2 3, , , , , .b a a c a d If wefix the letters “b”, “c” and “d”, the 3 different “a” s can be permuted amongst themselves in3! 6= ways. We now list down all the 6 permutations so generated on the left hand side in thefigure below, and see that these 6 permutations correspond to a single permutation if the “a” swere all the same:

1 2 3

1 3 2

2 1 3

2 3 1

3 1 2

3 2 1

b a a c a d

b a a c a d

b a a c a db a a c a d

b a a c a d

b a a c a d

b a a c a d

The 6 permutations on the left with different “a”s as correspond toa single permutation with the “a” s all same.

Fig - 6

Thus, the actual number of permutation with the “a”s all same will be

6! 6! 1203! 6

= =

(b) We now have 3 repeated “a” s and 2 repeated “b” s, and a total of 8 letters. If we for amoment take the “a” s and “b” s as all different, the total number of permutations of this set of8 letters would be 8

8 8!P =



However, once we list down these 8! permutations, we will see that (as in the previous part)3! = 6 permutations in this list will correspond to a single permutation if the “a”s were all thesame. Similarly, 2! = 2 permutations in this list will correspond to a single permutation if boththe “b”s were the same. Thus, the actual number of permutations if “a”s and “b”s were the

same would be 8!

3!2!(c) These results can now easily be generalised for this general set and the number of permutations

will be

( .....)!! ! !......

m n pm n p+ + +

____________________________________________________________________________________

From this example once again, the power of the fundamental principle of counting should be quite evident.Using a logical development/extension of this principle, we see that we’ve been able to solve non-trivialquestions like the one above.

Example – 10

Consider the integral equation 1 2 4x x+ = where 1 2,x x ∈ & . The non-negative solutions to this equationcan be listed down as { }0, 4 , { }1,3 , { }2, 2 , { }3,1 and { }4,0 . Thus, 5 non-negative integral solutionsexist for this equation.

We would like to solve the general case. How many non-negative, integral solutions exist for the equation

1 2 ..... nx x x r+ + + =

Solution: You might be surprised to know that this question can be solved using the general result obtainedin the previous example. Can you think how?Let us consider an arbitrary integral equation, say 1 2 3 8.x x x+ + = Consider any particular non-negative integral solution to this equation, say { }2,3,3 . We some how need to “tag” this solutionin a new form; a form which is easily countable. This is how we do it. We break up the solution2 + 3 + 3 = 8 as shown below:

11 111 111 8= ...(1)

Similarly, 1 + 6 + 1=8 would be written as

1 111111 1 8= ...(2)

and 0 + 1 + 7 = 8 would be written as

1 1111111 8= ...(3)

and 0 + 0 + 8 = 8 would be written as

11111111 8= ...(4)



An alert reader must have realised the ‘trick’ by now. In each of (1), (2), (3) and (4), we have onthe left hand side 8 “1” symbols and 2 “ ” symbols, in different orders. Any non-negative integral

solution can thus be represented by a unique permutation of 8 “1” symbols and 2 “ ” symbols.

Conversely, every permutation of 8 “1” symbols and 2 “ ” symbols represents a unique non-negative integral solution to the equation .

Thus, the set of non-negative integral solutions to the equation and the set of permutations of 8 “1”symbols and 2 “ ” symbols are in one-to-one-correspondence. To count the required number

of solutions, we simply count the permutations of 8 “1” symbols and 2 “ ” symbols, which as

described in the last example would be (8 2)! 10! 45

8!2! 8!2!+ = =

This beautiful artifice described about should make it clear to you the significance of (and thechallenge of producing!) elegant proofs/solutions.

We now generalise this result. Any non-negative integral solutions to the equation

1 2 ..... nx x x r+ + + = can be represented using r “1” symbols and n – 1 “ ” symbols. The totalnumber of permutation of these symbols will be

1( 1)!!( 1)!

n rr

n r Cr n

+ −+ − =−

and hence, this is the required number of solutions.

Example – 11

Consider a rectangular integral grid of size .m n× Forexample, a 4 5× integral grid is drawn alongside:A person has to travel from point A(0, 0) to the diagonallyopposite point C(m, n). He moves one step at a time, towardsthe east or towards the north (that is, never moves towardsthe west or south at any time). How many distinct paths existfrom the point A to the point C?

A B

CD

Fig - 7

Solution: Let us draw a random path on our 4 5× grid in Fig. 6 and think of some way to mathematicallyspecify/describe this path

A B

CD

Fig - 8

A random path across the 4 × 5 grid that our travelling person can follow.The question now is: How do we mathematically characterise this path?



Suppose you had to describe this path to a blind person. If you use E for a step towards the eastand N for a step towards the north, you’d tell the blind person that the travelling person took thefollowing path.

“ E E N E N E E N N ”

This string that we just formed should immediately make you realise how to calculate the numberall the possible paths. We have 5 “E” steps and 4 “N” steps. Any permutation of these 9 stepsgives rise to a different unique path. For example, the string “ E E E E E N N N N ” is the paththat goes straight east from A to B and then straight north from B to C. Thus, any path can beuniquely characterised by a permutation of these 9 steps. The number of permutations of these 9

letters, 5 of which are “E”s and 4 are “N”s, is 9!

5!4! . This is therefore the number of different paths

that the travelling person can take from A to C. For an m n× grid we will have (m + n) total steps,m of them being “E” s and the remaining n being “N” s Thus, the number of possible paths

is ( )!! !

m nm n

+×

.

Example – 12

(a) We have m apples, n oranges and p bananas. In how many ways can we make a non-zero selectionof fruit from this assortment?

(b) How many factors does 144000 have? In general, how many factors does a natural number N have?

Solution: These two seemingly unrelated questions have exactly the same approach to their solutions!Before reading the solution, can you imagine how?

(a) The most important point to realise in this question is the nature of objects to be selected. Wehave m apples. These m apples are exactly identical to each other. You cannot make out oneapple from another. This means that if you have to choose r apples out of n, there’s only oneway of doing it: you just pick (any of the) r apples. It doesn’t matter which apples youchoose, because all the apples are identical. Thus, only 1 r-selection is possible. So how manytotal selections are possible? We either select 0 apples, 1 apple, 2 apples, .... r apples, ... orm apples. Thus (m + 1) ways exist to select apples.

Make sure you properly understand the essence of this discussion and why

choosing apples out of can be done in only 1 and not ways.nrr n C

We therefore have (m + 1) ways to select apples, (n +1) ways to select oranges and (p + 1)ways to select bananas. The selection of a particular fruit is independent of the selection ofanother fruit.

Hence, we have (m +1) (n +1) (p +1) ways to select a group of fruit.



But wait! In the (m +1) ways of selecting apples, there’s also one way in which we select noapple. Similarly, in the (n + 1) ways of selecting oranges, there’s one in which we select noorange, and in the (p + 1) ways of selecting bananas, there’s one in which we select nobanana. Thus, in the product (m +1) (n +1) (p +1), there’ll be one case involving no fruit ofany type. We have to exclude this case if we want a non-zero selection of fruit.Therefore, the number of ways of making a non-zero selection is (m +1)(n +1)(p +1) —1.

(b) You might be wondering how the factors problem is related to the first part! Read on.Lets consider a smaller number first. Take 60, for example, and list down all its factors(including 1 and 60):

{ }1,2,3,4,5,6,10,12,15, 20,30,60

From elementary mathematics, you know that any number can be factorized into a product ofprimes. For example, 60 can be written in its prime factorization from as:

60 2 2 3 5= × × ×

2 1 12 3 5= ⋅ ⋅

Such a representation exists for every natural number N. Can we somehow use thisrepresentation to find the number of factors?

Consider any factor of 60, say 12, The prime factorization from of 12 is 2 12 3⋅ . Similarly, this

representation for 15, for example, is 1 13 5⋅ and for 30 would be 1 1 12 3 5⋅ ⋅ .With discussion in our mind, we rephrase our problem: We have 60 whose prime representationis 2 1 12 3 5⋅ ⋅ . Thus, we have 2 twos, 1 three and 1 five with us. (You could imagine that wehave 2 apples, 1 orange and 1 banana).To form a factor of 60, what we have to do is to make a selection of prime factors fromamongst the available prime factors. For 60, we have 2 twos (apples), 1 three (orange) and 1five(banana). In how many ways can we make our selection?

The last part tells is that we can do it in ( 1)( 1)( 1)m n p+ + + ways or (2 1)(1 1)(1 1) 12+ + + =in this particular case. (We also allow no selection of any prime factor - this corresponds to thefactor 1 of 60)Do you feel the elegance of this solution?For the general case of a natural number N whose prime representation is of theform 1 2

1 2 ........ ,nnp p pαα α the number of factors (including 1 and N) is

1 2( 1)( 1)........( 1)nα α α+ + +

For example, 144000 can be represented as7 2 3144000 2 3 5= ⋅ ⋅

The required number of factors is

(7 1)(2 1)(3 1) 96+ + + =



Example – 13

12 person are sitting in a row. In how many ways can we select 4 persons out of this group such that no twoare sitting adjacent in the row?

Solution: To make the question more clear, here’s a valid and a non-valid selection:

Valid

Non-valid

Fig - 9

We have to find some (mathematical) way of specifying a valid-selection. An obvious method thatstrikes is this: In our row, we represent every unselected person by U and every selected personby S. Thus, the valid selection in the figure above becomes:

U S U U S UU S U S UUTo count the numbers of valid selections, we count the number of permutation of this string above,consisting of 8 “U” symbols and 4 “S” symbols, subject to the constraint that no two “S” symbolsare adjacent.

To count such permutation, we first fix the 8 “U” symbols. There will then be 9 blank spacesgenerated for the “S” symbols as shown below:

__U__U__U__U__U__U__U__U__From these 9 blank spaces, any 4 spaces can be chosen and the “S” symbols can be put there andit’ll be guaranteed that no two “S” symbols will be adjacent. Thus, our task now reduces to justselecting 4 blank spaces out of the 9 available to us, which can be done simply in 9

4C ways.

These are the required number of selections.



Example – 14 DIVISION INTO GROUPS

We will use a standard pack of 52 cards to illustrate the concepts involved.

(a) (i) In how many ways can a pack of 52 cards be divided equally into 4 sets?(ii) Equally among 4 players?

(b) (i) In how many ways can this pack be divided into 4 sets of 9, 13, 14 and 16 cards each?(ii) Distributed among 4 players with 9, 13, 14 and 16 cards?

(c) (i) In how many ways can this pack be divided into 4 sets of 10, 14, 14 and 14 cards each?(ii) Distributed among 4 players with 10, 14, 14 and 14 cards?(iii) Distributed among 4 players with 12, 12, 14 and 14 cards?

Solution: For this question, you must understand that the division of a set into a number of subsets is differentfrom the division of a set into a number of subsets and then distribution of those subsets to differentpersons. For example, suppose you are alone by yourself and you are dividing your deck ofcards into 4 sets. What you’ll do is do the division of the deck and keep the 4 sets in front of you.Which set lies where in front of you doesn’t matter. Your only objective was to divide the pack into4 sets which you did. Precisely speaking, the order of the groups doesn’t matter.On the other hand, suppose you were playing a game of Bluff with 3 other people. Before startingthe game, you divide the deck into 4 sets of 13 cards each. But now you have to give each set of13 cards to a different person. Where a set of cards goes matters. In other words, the order of thegroups matters.As another example, suppose you had to divide a class into 3 sub-groups. You do your job-theorder of the groups doesn’t matter, you just have to divide them. But suppose you had to dividethe class into 3 sub-groups and place each sub-group in a different bogey of a roller coaster. Inthis case the order of the groups matters because which sub-group sits in which bogey matters.Once again, make sure that you understand very clearly when the order of the groups matters andwhen it doesn’t

(a) (i) We want to divide the pack into 4 equal sets of 13 cards each. We proceed as follows;

First select a group of 13 cards from the 52 cards; call this group A. This can be donein 52

13C ways. Group B of another 13 cards can now be selected from the remaining 39cards in 39

13C ways. Group C of another 13 cards from the remaining 26 can now beselected in 26

13C ways. This automatically leaves group D of the last 13 cards.Thus, the division can be carried out in

52 39 2613 13 13C C C× × ways.

However, notice that the order of the groups doesn’t matter, as discussed earlier. Thismeans that even if our order of group selection was BACD or ADBC (or for that matterany other permutation of ABCD), such a selection would essentially be the same as ABCD.

The number of permutations of ABCD is 4! = 24.



Thus, the actual number of ways of division is

( )52 39 2613 13 13

124

C C C× ×

1 52! 39! 26!24 13! 39! 13! 26! 13!13!

= × ×

( )452!

13! 4!=

⋅

(ii) If we had to distribute the cards equally among 4 players, the order of group selectionwould have mattered. For example, ABCD would be different from BACD. Thus, we

don’t divide by 4! in this case. The number of possible distributions is ( )452!13!

(b) (i) For this part, observe that the group sizes are all unequal. This is a somewhat differentsituation than the previous part where the group sizes were equal. We’ll soon see why.

We follow a similar sequence of steps as described earlier.

(i) Select group A of 9 cards from the deck of 52 : 529C ways

(ii) Select group B of 13 cards from the remaining 43 : 4313C ways

(iii) Select group C of 14 cards from the remaining 30 : 3014C ways.

(iv) This automatically leaves group D of the remaining 16 cards.The number of ways for this division is

52 43 309 13 14C C C× ×

52! 43! 30!9! 42! 13! 30! 14!16!

= × ×

52!

9!13!14!16!=

From now on, note and remember that

( )! ! ! !

m n p q n p q p q qm n p qC C C C

m n p qm n p q

+ + + + + +

× × ×

+ + + =

Why don’t we divide by 4! here as we did earlier, even when the order of groupsdoesn’t matter? This is because the group sizes are unequal. A particular selection in aparticular order, say ABCD, will never be repeated in any other order of selection.Our order of selection (in terms of group size) is 9, 13, 14 and 16 cards. Carefullythink about it; a particular selection like ABCD will be done only once (and not 4!times as in the previous part due to equal group sizes)



(ii) If we had to distribute the 4 sets to 4 players, the situation changes because which subsetof cards goes to which player matters. A particular group of sets of cards, say ABCD, canbe distributed among the 4 players in 4! ways. Thus, now the number of distributions willbe

no. of ways of division of the no. of ways of Distribution deck as specified into the 4 sets of those sets to the 4 players

×

( )52! 4!9! 13! 14! 16!

= ×

(c) (i) In this part, the situation is a hybrid of the previous two parts. 3 of the 4 groups are equalin size while the 4th is different.

We first select 10 cards out of 52. This can be done in 5210C ways. The remaining 42

cards can be divided into 3 equal groups (by the logic of part -(a)) in

342! 1

3!(14!)×

(Division by 3! is required since the order of groups doesn’t matter. Thus, a particularorder of 15-card groups, say ABC, is, for example, the same as ACB)Thus, the required number of ways is

5210 3

42! 13!(14!)

C × ×

352! 1

3!10!(14!)= × ...(1)

(ii) For the second question where we have 4 players and we want to give them 10, 14, 14and 14 cards, we already have obtained the number of possible division in (1). To distributethe subsets, each division of 4 sets can be given to the 4 players in 4! ways. Thus, thenumber of distributions of the 4 subsets is

( )3 352! 1 4 52!4!

3!10!(14!) 10! 14!×= × × =

We can also look at this current problem in the following way. We want to distribute the52 cards to the 4 players in sets of 10, 14, 14 and 14.We first decide which guy to give the set of 10 cards. This can be done in 4 ways. Nowwe choose 10 cards for him (in possibly 52

10C ways). The other three players will nowget the remaining 42 cards equally. We can select 12 cards for one of them in 42

14C ways.The third one can get another 14 cards in 28

14C ways and finally the fourth one gets theremaining 14 cards.Thus, the number of possible ways of distribution is 52 42 28

10 14 144 C C C× × ×

352!

10! (14!)4×=



(iii) Finally, we now have two sets of one size and two of another size. We first just carry outthe division into subsets, without assigning any subset to any player.

This can be done in

52 40 28 1412 12 14 14

Group A of Group B of Group C of Group D of Groups A and Groups C and12 cards 12 cards 14 cards 14 cardsB are of the D are of the same sizer same sizer

1 12! 2!

C C C C↓ ↓

× × × × ×! ! !

( )2 2 252! 1

(12!) (14!) 2!= ×

Once the division of the deck into the 4 subsets has been accomplished, we assign asubset to each of the 4 players. This can be done in 4! ways.

Therefore, the possible number of ways of distribution of the cards to the 4 players is

( )2 2 252! 1 4!

(12!) (14!) 2!× ×

Example – 15 CIRCULAR PERMUTATIONS

In this example, we talk about circular permutations, i.e., arrangements in a circular fashion(a) In how many ways can n people be seated around a circular table?(b) We have a group of 5 men and 5 women. In how many ways can we seat this group around a

circular table such that:(i) all the 5 women sit together (ii) no two women sit together.

(c) In how many ways can a necklace be formed from n different beads?Solution: Circular permutations are somewhat different from linear permutations. Lets see why.

Consider the letters A, B, C and D. The 4 linear permutations ABCD, BCDA, CDAB and DABCcorrespond to a single circular permutation as shown in the figure below:

In a circular permutation, only the relative ordering of the symbols matters.The 4 linear permutations ,

and all have the same relative ordering so that they correspond to a single circular permutation.

ABCD BCDA, CDAB DABC

Fig - 10

A

B

C

D

This means that the number of circular permutations of ABCD is only 4! 3! 64

= =



(a) We have n people, say 1 2, ............ .nP P P As just described, n linear permutations of thesepeople will correspond to a single circular permutation as depicted in the figure below:

1 2

2 3 1

3 4 1 21 2

1 2 1

............................................. ...........::::...............

linear permutations a single circular permutationcorrespond to because of

n

n

n

n n n

PP PP P P PP P PP PP P

P P P P

n

− −

→ the same relative ordering.

Thus, the number of circular permutations is ! ( 1)!n nn

= −

We can also arrive at this number in another way. We take a particular person, say 1P , andseat him anywhere on the table. Once 1 'P s seat becomes fixed, the rest of the (n – 1) seatsbear a fixed relation to 1 'P s . In other words, once 1 'P s seat becomes fixed, we can treat the(n – 1) seats left as a linear row of (n –1) seats. Thus, the remaining (n –1) people can beseated in (n –1)! ways.

(b) Let the 5 women be represented by 1 2 3 4 5, , , , and W W W W W and the 5 men by

1 2 3 4 5, , , , and M M M M M . Since we need all the women to sit together, we first treat all the5 women as a single entity W. Now, the 5 men and the entity W can be seated around the tablein ( )(5 1) 1 !+ − ways, i.e., 120 ways.

(i) Once the 5 men and the entity W have been seated, we now permute the 5 women insidethe entity W. This can be done in 5! = 120 ways. The total number of ways is thus120 × 120 = 14400

M1

W1

M3

M4

M5

First seat the 5 menand the entity W

Then permute the womeninside the entity .W

(i)

(ii)

Fig - 11

W2 W3 W4 W

5The entity W

M2



(ii) Since we want no two women to sit together, we first seat all the 5 men, which can bedone in (5 –1)! = 24 ways. Seating the 5 men first creates 5 non-adjacent seats where thewomen can then be seated in 5! = 120 ways.

M1

M2

M3M4

M5

First seat the 5 men,4! ways are possible

Then seat the 5 womenin the 5 spaces createdas shown in the figure5! ways are possible

(i)

(ii)

Fig - 12

The total number of ways is this 5! × 4!.

(c) The situation of a necklace is slightly different than that of seating people around a circulartable. The reason is as follows:

Suppose we have 5 beads A, B, C, D and E. Consider two circular necklaces of these 5beads

A A

B BEE

C CDD

Necklace-1 Necklace-2

Fig - 13

Are these two necklaces different? No, because a-necklace can be worn from both ways.Necklace-2 is the same as necklace-1 if I look into it from the other side of the page. In otherwords, for a necklace, a clockwise permutation and its corresponding anti-clockwisepermutation are identical. Thus, the number of circular permutations would reduce by a factor

of two, i.e., the number of different necklaces possible is 1 ( 1)!.2

n −



TRY YOURSELF - I

Q. 1 How many words can be made with the letters of the word “TECHNOLOGY” which do not beginwith T but end with Y ?

Q. 2 In how many ways can the letters of the word CINEMA be arranged so that the order of vowels donot change?

Q. 3 How many sides are there in a polygon which has 35 diagonals?

Q. 4 In how many ways can 6 boys and 4 girls sit in a row so that no boy is between two girls?

Q. 5 How many words can be made with the letters of the word BHARATI so that all the vowels areconsecutive?

Q. 6 Let ∈n N and 300 < n < 3000. If n is made of distinct digits by taking from 0, 1, 2, 3, 4, 5 then findthe greatest possible number of values of n.

Q. 7 How many words can be made with the letters of the word INSTITUTION such that vowels andconsonants alternate?

Q. 8 How many different committees of 5 members can be formed from 6 men and 4 ladies if eachcommittee is to contain at least one lady?

Q. 9 Six Xs are to be placed in the squares of the given figure, such that each row contains at least one X.In how many different ways can this be done?

Q. 10 In how many ways can 16 identical mangoes be distributed among 4 persons if none gets less than 3mangoes?



Section - 5 Section - 1MORE APPLICATIONS

By now, you should have a pretty good idea about the basics of permutations and combinations. In thissection, we will encounter more advanced problems.All the questions discussed in the following pages are directly or indirectly based on the concepts alreadydiscussed in the previous sections. In case you find anything confusing, refer to the relevant parts again.

Example – 16

On a standard 8 × 8 chessboard, find the(a) number of rectangles (b) number of squares

Solution: (a) Visualise any arbitrary rectangle on the chessboard, say the one depicted on the left in thefigure below:

P

Q

An arbitrary rectangle on thechessboard. How to specifythis rectangle is explainedin the figure on the right

To mathematically characterise the rectangle that we selected, we specify the pair of vertical edges

and and the pair of horizontal edges and . Doing so uniquely determines the rectangle X Y P Q

X Y

As explained in the figure above, any rectangle that we select can be uniquely determined byspecifying the pair of lines X and Y that make up the vertical edges of the rectangle and the pairof lines P and Q that make up the horizontal edges of the rectangle.

On the chessboard, there are 9 vertical lines available to us from which we have to select 2.This can be done in 9

2C ways. Similarly, 2 horizontal lines can be selected in 92C ways. Thus,

the total number of rectangles that can be formed is 9 92 2 1296C C× = .

(b) To select a square, observe that the pair of lines X and Y must have the same spacing within Xand Y as the pair of lines P and Q. Only then can the horizontal and vertical edges of theselected rectangle be of equal length (and thus, the selected rectangle is actually a square).



In how many ways can we select a pair (X, Y) of lines which are spaced a unit distance apart? Its obviously 8. Corresponding to each of these 8 pairs, we can select a pair (P, Q) of linesin 8 ways such that P and Q are a unit distance apart. Thus, the total number of unit squares is8 × 8 = 64 (This is obvious otherwise also). Now we count the number of 2 × 2 squares. Inhow many ways can we select a pair (X, Y) of lines which are 2 units apart ? A little thoughtshows that it will be 7. Corresponding to each of these 7 pairs, we can select a pair (P, Q) oflines in 7 ways such that P and Q are 2 units apart. Thus, the total number of 2 × 2 squaresis 7 × 7 = 49.

Reasoning this way, we find that the total number of 3 × 3 squares will be 6 × 6 = 36, the totalnumber of 4 × 4 squares will be 5 × 5 = 25 and so on.

Thus, the total number of all possible squares is

7 7 8 81 1 squares 2 2 3 3squares squaressquares squares

64 49 36 ........... 4 1 204↑ ↑↑ ↑ ↑× ×× × ×

+ + + + + =

Example – 17

Consider an n-sided convex polygon.(a) In how many ways can a quadrilateral be formed by joining the vertices of the polygon ?(b) How many diagonals can be formed in the polygon?

Solution: (a) Observe that a selection of any 4 points out of the n vertices of the quadrilateral will give riseto a unique quadrilateral (since the polygon is convex, the problem of our selection containingall or 3 collinear points does not exist). 4 points out of n can be selected in 4

nC ways. Thus,we can have 4

nC different quadrilaterals.

(b) To form a diagonal, we need 2 non-adjacent vertices ( because 2 adjacent vertices will forma side of the polygon and not a diagonal). The total number of ways of selecting 2 vertices outif n is 2

nC . This number also contains the selections where the 2 vertices are adjacent. Thoseselections are simply n in number because the polygon has n sides. Thus, the total number ofdiagonals is

2nC n−

( 1)2

n n n−= −

( 3)2

n n −=

Example – 18

Give a combinatorial (logical) justification for this assertion:1 2 1

0 1 2 ...n n n n r n rr rC C C C C+ + + + ++ + + + =



Solution: The right hand side tells us that we have to select r persons out of a group of 1n r+ + persons. Todo so, we consider any particular group of r persons from these 1n r+ + persons. Specify theser persons by the symbols 1 2, ... rA A A .

Now, to count all the possible r-groups from this group of 1n r+ + , we consider the followingmutually exclusive cases:

(1) The r -group does not contain A1

Such r-groups can be formed in n rrC+ ways since we have to select r people out of n + r.

(2) The r -group contains A1 but not A2

We have to select (r – 1) people out of ( 1)n r+ − because we already have selected 1Aso we need only r –1 more people and since we are not taking 2A , we have ( 1)n r+ −

people to choose from. This can be done 11

n rrC+ −− ways.

(3) The r -group contains A1, A2 but not A3

We now have to select (r – 2) people out of ( 2)n r+ − . This can happen in 22

n rrC+ −−

ways. Proceeding in this way, we arrive at the last two possible cases. '

(r) The r -group contains A1, A2 ... Ar – 1 but not Ar .We need to select only 1 person out of (n + 1) available for selection. This can be done in

11

n C+ ways.

(r + 1)The r-group contains A1, A2 ... ArIn this case, our r-group is already complete. We need not select any more person. Thiscan be done in 0

nC or equivalently 1 way.

Convince yourself that these (r + 1) cases cover all the possible cases that can arise in the formationof the r - groups. Also, all these cases are mutually exclusive. Thus, adding the number of possibilitiesof each case will give us the total number of r-groups possible, i.e.

1 2 1 10 1 2 1...n n n n r n r n r

r r rC C C C C C+ + + − + + +−+ + + + + =

Example – 19

How many distinct throws are possible with a throw of n dice which are identical to each other, i.e. indistinguishableamong themselves ?

Solution: The important point to be realised here is that the dice are totally identical. Suppose we had just2 dice, say die A and die B. Suppose that, upon throwing these dice, we get a “two” on A and a“three” on B. This case would be the same as the one where we get a “three” on A and a “two”on B because we cannot distinguish between A and B. What we are concerned with is only whatnumbers show up on the top of the dice. We are not concerned with which die shows whatnumber. This means that if we have n dice and we throw them, we are only concerned with howmany “ones”, “twos”, “threes” etc show on the top faces of the dice; we are not at all interested inwhich die throws up what number.



If we denote the number of “ones” we get by 1x , number of “twos” we get by 2x and so on, wewill have

1 2 6...x x x n+ + + =

Thus, the total number of distinct throws will be simply the number of non-negative solutions to thisintegral equation.

As discussed earlier, this number will be 6 1 56 1 5

n nC C+ − +− = .

What would be the number of distinct throws if the n dice were not identical?

Example – 20

Give combinatorial arguments to prove that

1

1

2n

n nr

rr C n −

=

⋅ = ⋅∑

Solution: Let us first interpret what the left hand side of this assertion says.

Suppose we have a group of n people. We select a sub-group of size r from the group of npeople. This can be done in n

rC ways. Once the sub-group has been formed, we select a leaderof that sub-group, and send that sub-group on an excursion. The leader can be selected in r ways.Thus, the total number of different ways in which an r - group can be formed with a unique leaderis n

rr C× .

Now r can take any integer value from 1 to n, i.e. 1 r n≤ ≤ . Thus, the total number of all possible

sub-groups, each sub-group being assigned a unique leader, will be 0

nn

rr

r C=

⋅∑ which is the left

hand side of our assertion.

To prove this equal to the right hand side, we count the sub-groups from a different angle. Wecount all those sub-groups in which a particular person, say A, is the leader.

Since A is the leader, A is fixed in our sub-group. For each of the remaining (n – 1) people, wehave two options. We either put the person in the group led by A or we don’t. Thus, the totalnumber of sub-groups in which A is the leader will be

1

( 1) times

2 2 2 ... 2 2n

n

−

−

× × × × ="##$##%

Since any of the n persons can be the leader, and under each person’s leadership, 12n− groups canbe formed, the total number of sub-groups, each sub-groups under some unique person’s leadershipis 12nn −⋅ . This proves our assertion that

1

1

2n

n nr

rr C n −

=

⋅ = ⋅∑



Example – 21

A composition of a natural number N is a sequence of non-zero integers {a1, a2 .......... ak} which add up to N.How many compositions of N exist?

Solution: Let us make the question more clear by taking a particular example for N, say N = 4.

As described in the question, the compositions of N = 4 will be

{ } { } { } { } { }4 , 1, 3 , 2,2 , 3,1 , 1,1, 2 , { } { } { }1,2,1 , 2,1,1 and 1,1,1,1 which are 8 in number.

Observe carefully the compositions listed out. How can we characterize each of these compositionsmathematically? Recall the problem of finding the number of non-negative solutions of the integralequation 1 2 .......+ + + =nx x x r where each solution corresponded to a unique permutation of r

“ 1 “symbols and n - 1 " "symbols. Can we do something like that here? In other words, can werepresent each composition in another form whose permutations are easier to count?

It turns out that we can, as follows:

{4} = 1 1 1 1

{1, 3} = 1 + 1 1 1

{3, 1} = 1 1 1 + 1

{2, 2} = 1 1 + 1 1

{1, 1, 2} = 1 + 1 + 1 1

{1, 2, 1} = 1 + 1 1 + 1

{2, 1, 1} = 1 1 + 1 + 1

{1, 1, 1, 1} = 1 + 1 + 1 + 1

On the right hand side, we have 4 “1”s and thus 3 blank spaces between the 4 “1” s. We can insert“+” signs in these blank spaces; each different arrangement of “+” signs in these blank spaces willcorrespond to a different composition.

To count the number of these arrangements, we proceed as follows: For each blank space, wehave 2 options, we can either insert or not insert a “+” sign into that space. There are 3 blankspaces; so the total number of all arrangements of “+” signs in the 3 blank spaces is 2 × 2 × 2 = 8(which is the number of compositions we already listed out).

In the general case, we will have (N – 1) blank spaces and 2 options for each such blank space.Thus, the total number of ways in which we can arrange “+” signs in these blank spaces, and

therefore, the total number of compositions, will be ( )

1

1 times

2 2 2 ........2 2 −

−

× × × ="##$##%N

N



Example – 22

Prove logically the following assertion:

0

2 3n

k n nk

kC

=

⋅ =∑Solution: Suppose that we have to form a string of length n, consisting of only letters from the set {A, B, C}.

Thus, we have 3 options to fill any particular place in the string: We fill that place with either A, Bor C. Thus the total number of different strings of length n would be

times

3 3 3 ........ 3 3× × × × ="##$##%n

n

We now approach the task of formation of these strings from a different perspective.Suppose that our string contains a total of r “A” s. How many such strings will exist? We firstselect r places out of n which we will fill with “A”. This can be done in nCr ways. For each of theremaining (n – r) places, we have two options; we either fill it with “B” or “C”. Thus, the numberof strings containing r “A” s will be nCr .2

n–r.Now we vary r from 0 to n and thus get the total number of strings as

( )0 0

2 2 wheren n

n n r n kr n k

r kC C k n r−

−= =

⋅ = ⋅ = −∑ ∑

0

2n

n kk

kC

== ⋅∑ ( )since n n

n k kC C− =

Thus, 0

2 3n

n k nk

kC

=

⋅ =∑

Example – 23

Prove logically the following assertion:

1 2 11......n n n r n

r r r r rC C C C C− − +++ + + + =

Solution: The right hand side says that we have to select (r + 1) people out of a group of (n +1).

To do so, we list down the following (mutually exclusive) cases which exhaust all the possiblecases:

(1) The group contains A1 : nrC ways

(since we have to select r people moreapart from A1 from n that are availablefor selection)



(2) The group does not contain : 1−nrC ways

A1 but contains A2 (since we have to select r people moreapart from A2 from (n – 1) that areavailable for selection)

(3) The group does not contain A1 and A2 : 2−nrC ways

but contains A3 (since we have to select r people moreapart from A3 from (n – 2) that areavailable for selection)

'

(n – r) The group does not contain : 1rrC+ ways

A1, A2 ....... An–r–1 but contains An–r (since we have to select r peoplemore apart from An–r from the (r + 1)that are available for selection

(n – r + 1) The group does not contain : rrC ways (since we have to select r

1 2, ... n rA A A − but contains 1n rA − + more people apart from 1n rr − + from theremaining r that are available forselection)

Convince yourself that all these cases are mutually exclusive and exhaust all the possibilities. Thus,the total number of (r + 1) - groups from (n + 1) people is

1 1 21 ...n n n n r

r r r r rC C C C C+ − −+ = + + + +

Example – 24

Consider n points in a plane such that no three of them are collinear. These n points are joined in all possibleways by straight lines. Of these straight lines so formed, no two are parallel and no three are concurrent. Findthe number of the points of intersection of these lines exclusive of the original n points.Solution: To gain more insight into the problem, let us consider the case when n = 4.

Fig - 15

AB 1

C

3

D

2



We originally have 4 points in the plane labelled as A, B, C and D. When we draw all the straightlines possible (by joining every possible pair of points), we see that 3 new intersection points aregenerated, labelled as 1, 2 and 3.As we now discuss the general case, refer to this figure for help.Since we have n points, the number of straight lines that can be generated is equal to the numberof pairs of points that can be selected from n points, which is equal to 2

nC . Every pair of straightlines so generated will intersect at some point. Thus, the total number of intersection points if wecount this way should be equal to the number of pairs of straight lines possible from the 2

nC lines,which is 2( )

2

nC C .However, observe that in this number, the original intersection points have also been counted, andthat too multiple times. Lets determine how many times a particular point, say A has been counted.(n – 1) lines pass through A. Thus 1

2n C− pairs of straight lines are possible from these (n – 1) lines.

Thus, A has been counted 12

n C− times. This means that the n original points have been counted1

2nn C−× times.

To calculate the new intersection points, we subtract this number from the (supposed) number ofintersection points we obtained earlier.Thus, the actual number of new intersection points is

2( ) 12 2

nC nC n C−− ×

( 1) ( 1) 1( 1)( 2)2 2

2 2

n n n nn n n

− − − × − − = −

2( 1)( 2) ( 1)( 2)8 2

n n n n n n n− − − − −= −

2( 1){( 2) 4( 2)}8

n n n n n− − − − −=

2( 1)( 5 6)8

n n n n− − +=

( 1)( 2)( 3)8

n n n n− − −=

Verify that this formula works for n = 4

Example – 25

How many 4 letters strings can be formed from the letters of the work INEFFECTIVE ?Solution: Observe that some letters repeat more than once:

Letter E F I C T V N

Frequency 3 2 2 1 1 1 1

This means that our string of 4 letters could contain repeated letters. (It’s thus obvious that wecannot straightway use n

rP to evaluate the number of strings.



In such a case, we divide the types of strings that we can form into different mutually exclusivecases:

Case 1: All 4 letters are different

In this case, we have 7 letters to choose from and we have to arrange them in 4places. The number of such strings will be 7

4 840P =

Case 2: 2 letters are same, the other 2 are different

We have 3 types of letters (E, F and I) that can be repeated twice. Thus, the twice-repeated letter can be selected in 3 ways. Once that is done, the rest of the 2 letterscan be selected from amongst the remaining 6 options in 6

2 15C = ways. Once we’

ve formed the combination of the 4 letters we can permute them in 4! 122!

= ways. The“2!” occurs in the denominator because of the twice-repeated letter.

Thus, the total number of such strings will be 3 15 12 540× × = .

Case 3: 2 letters are the same, the other 2 are also the same

For example, the string “EFEF” will be such a string.There are 2 letters now that we want to be twice-repeated; observe that there areonly 3 types of letters (E, F and I) that can be twice repeated. Thus, the letters that willoccur in the string can be selected in 3

2 3C = ways. Once that is done, we can

permute the 4 letters in 4! 6

2!2!= ways.

The total number of such strings is therefore 3 × 6 = 18.

Case 4: 3 letters are the same, the 4th is differentOnly “E” can be repeated thrice so that E must occur in this string. The 4th letter can

be selected in 6 ways. Then, the combination so formed can be permuted in 4! 43!

=

ways.The number of such strings is 6 × 4 = 24.

Verify that these four cases are mutually exclusive and they exhaust all the possibilities.

The total number of strings is 840 + 540 + 18 + 24 = 1422

Example – 26

(a) In how many ways can 3 girls and 9 boys be seated in 2 vans, each having numbered seats with 3 seatsin the front and 4 seats at the back?

(b) How many seating arrangements are possible if 3 girls should sit together in a back row on adjacentseats?



Solution: (a) We have 12 people to seat and 14 available seats. We first select the 12 seats on which to seatthese people. This can be done in 14

12C ways.Once that is done, we can permute the 12 people amongst these 12 seats in 12! ways.Thus, the total number of seating arrangements is 14

12 12!C × .(b) Since the 3 girls need to be seated together in one back row, we first select the back row. This

can be done in 2 ways since there are 2 vans. In this row, 3 adjacent seats can be selected in2 ways.

selection of 3 adjacent seats can be done in 2 ways.

"#$#% "#$#%

In the 3 adjacent seats, the 3 girls can be permuted in 3! ways.

Thus, the girls can be seated in 2×2×3! = 24 ways. We now have 11 remaining seats. The 9boys can be seated in these 11 seats in 11

9P ways.

Thus, the total number of ways to seat the entire group is

119

11!24 24 12 11! 12!2!

P× = × = × = .

Example – 27

We have 3n objects, of which n are identical and the rest are all different. In how many ways can we select nobjects from this group ?

Solution: To select n objects, we select k objects from the identical ones and the remaining (n – k) from thedifferent ones. ( k will vary from 0 to n) As discussed somewhere earlier, from the group contain-ing identical objects, there will always be only 1 way of selection. From the group of 2n non-identical objects, (n – k) objects can be selected in 2n

n kC − ways.

Thus, the group of n objects containing k identical objects and the remaining (n – k) as non-identical objects can be formed in 2 21 n n

n k n kC C− −× = ways. The total number of ways is S where

2 2 2 2

1 00

...n

n n n nn k n n

kS C C C C− −

== = + +∑

A closed-form expression for S can be obtained by the following manipulation :

2 2 21 02 2( ... )n n n

n nS C C C−= + + +

2 2 21 0

2 2 21 0

...

...

n n nn n

n n nn n

C C C

C C C

−

−

+ + + = + + + +



2 2 21 0

2 2 21 2

...

...

n n nn n

n n nn n n

C C C

C C C

−

+

+ + + = + + + +

2 22

For all the terms in the lower row, we can use

n nr n rC C −

=

2 2 2 2 20 1 2 2( ... )n n n n n

n nC C C C C= + + + + +

2 22n nnC= + (This step uses the result of example - 7)

⇒ 2 2 11 22

n nnS C −= ⋅ +

Example – 28

5 balls are to be placed in 3 boxes. Each box can hold all the 5 balls. In how many ways can we place the ballsinto the boxes so that no box remains empty, if

(a) balls and boxes are all different

(b) balls are identical but boxes are different

(c) balls are different but boxes are identical

(d) balls as well as boxes are identical

(e) balls as well as boxes are identical, but the boxes are kept in a row.

Solution: One of the constraints that should always be satisfied is that no box should remain empty. Thus,each box should get at least one ball. This means that the distribution of balls can have theconfiguration ( 1, 1, 3) or (1, 2, 2). Only in these two configurations does no box remain empty.

Readers might observe some similarity between this problem and Example - 14 where we discussed division of a deck of cards into groups. Observe carefullywhich part in this problem corresponds to which case in Example - 14

(a) In this case, the balls and the boxes are all different. If you observe carefully, you will note thatthis case is equivalent to dealing cards to players. Here, we are dealing balls (all different) intoboxes (all different). In the card game, we were dealing cards (all different) to players(all different).Suppose we distribute the balls in the configuration (3, 1, 1). We first divide the group of ballsinto this configuration. This can be done in 5

3C ways since we just need to select a group of 3balls and our division will be accomplished. Once the division of the group of balls into 3-subgroups in the configuration (3, 1, 1) has been done, we can permute the 3 sub-groupsamong the 3 different boxes in 3! ways.

Thus, the number of ways to achieve the (3, 1, 1) configuration is 53 3! 60C × =

We now find the number of ways to achieve the (1, 2, 2) configuration .



We first select 2 balls out of the 5 which can be done in 52C ways. We then select 2 balls from

the remaining 3, which can be done in 32C ways. Thus simple division into the configuration

(2, 2, 1) can be achieved in 5 3

2 2

2!C C×

ways. Division by 2! is required since two subgroups

are of the same size and right now we are just dividing into sub groups so the order to the sub-groups doesn’t matter. Once division has been accomplished, we permute the 3 subgroups soformed amongst the 3 different boxes is 3! ways.

Thus, the number of ways to achieve the (2, 2, 1) configuration is 5 3

2 2 3! 902!

C C× × =

The total number of ways is 60 + 90 = 150To make things more clear, let us list down in detail the various configurations possible for the3 different boxes, A, B & C:

Box - A Box - B Box - C Number of ways1 1 3 201 3 1 203 1 1 201 2 2 302 1 2 302 2 1 30

Total:150

(b) The balls are now identical so it doesn’t matter which ball goes into which box. What mattersis only the configuration of the distribution.By simple enumeration, only 6 configurations exist for this case. (The notation [a, b, c] impliesthat Box - A has a balls, Box - B has b balls and Box - C has c balls.)

[ ] [ ] [ ]1,1,3 1,3,1 3,1,1

[ ] [ ] [ ]1,2, 2 2,1, 2 2, 2,1Thus, 6 possible ways exist for this case

(c) The boxes are identical. This means that it does not matter which sub-group of balls you put inwhich box. What matters is only the division of the group of balls. This case is akin to the onewhere you have to divide a deck of cards into sub-groups. (you aren’t required to distributethose sub-groups to players).

For the configuration (1, 1, 3), the number of ways of division is 53 10C = (we just choose 3

balls out the 5 and the division is automatically accomplished. For the configuration (1, 2, 2),

the number of ways of division is 5 3

2 2 152!

C C× = (division by 2! is required since two

sub-groups are of the same size, and here the order of the group doesn’t matter)Thus, the total number of ways is 10 + 15 = 25



(d) This case is quite straightforward. The balls are identical. The boxes are identical too. Theonly 2 possible configuration are (1, 1, 3) and (1, 2, 2). There can be no permutation of theseconfigurations since the boxes are indistinguishable. Thus, only 2 ways of division exist.

(e) If we keep the boxes in a row, we have inherently ordered them and made them non-identical,since the boxes can be numbered now.

Therefore, in this case, the balls are identical but the boxes are different so this questionbecomes the same as the one in part - (b)

Example – 29 PIRATES OF THE CARRIBEAN

In the Caribbean Sea, 13 pirates, while plundering an English ship, come upon a chest full of gold. Since thepirates have found the chest simultaneously, no one can claim the chest as his own.

To protect the chest from the avarices of the pirates, the pirate leader Captain Jack Sparrow suggests ascheme. “We must put a certain number of locks on this chest and distribute their keys amongst ourselves insuch a way that it can be opened only when 7 or more then 7 pirates decide to open it. In other words, onlywhen a majority of the 13 pirates agree should the chest be able to be opened.”

How can this scheme be implemented? What would be the minimum number of locks required and how musttheir keys be distributed?

Solution: This interesting problem can be easily solved by the application of the elementary conceptsdeveloped in this chapter.

Let us rephrase our problem slightly.We want that only a group of 7 or more pirates should be able to open the chest. This means thatwhenever such a majority group decides to open the chest, they should have amongst themselvesthe keys to all the locks on the chest.Suppose on the other hand, that only 6 pirates decide to open the chest. Then there should be atleast one lock on the chest whose key(s) are not with anyone amongst that group of 6. Thus, thatsingle lock will prevent the minority group of 6 pirates from opening the chest.This is the approach we now follow. For every possible group of 6 pirates, we put a lock on thechest and distribute 7 keys of the lock amongst the remaining 7 pirates. That lock will prevent ourgroup of 6 pirates from opening the chest.Such a lock will exist for every group of 6 pirates. Thus, whenever any group of 6 pirates decidesto open the chest, they will be prevented by one lock whose keys are with the other 7 pirates (thecase when even less than 6 pirates decide to open the chest is automatically solved because thenthere will be more than one lock to prevent that group from opening the chest)Also, whenever any group of 7 pirates decides to open the chest, there is no lock whose key isnot amongst one of the 7 members of that group. Thus, any group of 7 pirates (or more) will beable to open the chest.

Thus, what we need to do is put ( )13 136 7C C= locks on the chest and for each lock, we select a

group of 7 pirates, make 7 keys of that lock and give one key to each member of this group.



Example – 30

Prove that (n!)! is divisible by (n!)(n–1)!

Solution: We can equivalently show that ( )

( )( )1 !

! !

! n

n

n − is an integer.

To keep things easier, let us take n = 6. Thus, we need to show that ( )( )5!

6! !6!

is an integer. (You must

appreciate the magnitude of the numbers we are dealing with here!)Consider the following sequence of symbols:

1 1 1 1 1 1 2 2 2 2 2 2 120 120 120 120 120 120....................a a a a a a a a a a a a a a a a a a

This is a sequence of symbols whose length is 720, because each symbol ai occurs 6 times in thissequence and there are in total 120 symbols, and thus the total number of symbols is6 × 120 = 720. Let us find the number of permutations of this string. Since there are 120 ‘types’of symbols, each symbol being repeated 6 times, we will have to divide by 6! a total of 120 times,i.e, the number of permutations of this string will be

( )

120 times

720 !6! 6! 6! .......6!× × ×"###$###%

( )( )5!

6! !6!

=

Since the number of permutations of any string must obviously be an integer, the term ( )( )5!

6! !6!

=

must be an integer!We can now easily generalise this result.

Example – 31

n persons are seated around a circular table. In how many ways can we select 3 persons such that no 2 ofthem are adjacent to each other on the table?Solution: We will solve this problem by first considering the total number of selections possible and then

subtracting from this, the number of cases where 2 persons or all the 3 persons are adjacent.Observe the following figure for reference.

P1 P2

P3

P4

Pn

Pn-1

Fig - 16



The total number of ways of selecting 3 persons out of these n is 3.nC

Let us count the cases where only 2 of the 3 persons are adjacent. 2 adjacent people can beselected as follows:

{ } { } { } { }1 2 2 3 3 4 1or or ...............or nPP P P P P P P

Thus, there are n ways in which 2 adjacent persons can be selected. Once these 2 have beenselected (for example, say we select {P2 P3}), we can select the third non-adjacent person in(n – 4) ways (since for example, if we selected {P2 P3}, we cannot select P1 or P4). Thus thecases where only 2 of the 3 people are adjacent are n (n – 4) in number.We now count the cases where all 3 persons are adjacent. 3 persons can be adjacent in thefollowing manner:

{ } { } { }1 2 3 2 3 4 1 2or or ..................or nPP P P P P P PP

Thus, observe that there are n ways of selecting 3 persons who are adjacent.Finally, the number of ways to select 3 persons such that no 2 are adjacent is

( )3 4n C n n n− − −

( )( ) ( )1 2

46

n n nn n n

− −= − − −

( )( )1 4 56

n n n= − −

Example – 32

We have 21 identical balls available with us which we need to be distributed amongst 3 boys A, B and C suchthat A always gets an even number of balls. How many ways of doing so are possible?

Solution: The only constraint is that A should get an even number of balls. There’s no constraint on theminimum number of balls a boy should get. This means that a boy can also not be given any ball.

We can represent the number of balls given to A by 2x since A must get an even number of balls.If we represent the number of balls given to B and C by y and z, we should have

2 21x y z+ + = ... (1)

This means that to find the number of distributions possible, we find the number of non-negativeintegral solutions to the equation (1).

Note that x can take a maximum value of 10 and a minimum value of 0.

We rearrange (1) so that we get an integral equation with y and z as variables, treating x as aconstant

21 2y z x+ = − .... (2)



The number of non-negative integer solutions of (2) is

21 2 2 1 22 21 1 22 2x xC C x− + − −= = −

We now add the number of solutions so obtained for all the possible values of x. The total number

of solutions is therefore ( )10

0

22 2 132.x

x=

− =∑

Note: An alert reader must have noticed that we can form arbitrarily complex integer equations.For example, what do we do if we intend to find out the number of non-negative integersolutions to the equation 1 1 2 2 ........... n nx x x rα + α + α =

where the 'i sα are integers that are not necessarily equal to unity.

In addition, what if we impose constraints on the 'ix s themselves say, we define an

upper and lower bound for xi , i.e

i i il x u≤ ≤

For example, how do we distribute 20 apples among 4 boys such that each boy gets morethan 2 apples but less that 8 apples, in addition to the constraint that one of the boys, sayA, must always get an even number of apples? Think about it. We will revisit the problemof such general integral equations in Example - 39 and in the chapter on BinomialTheorem. There we’ll learn to solve such problems using the Multinomial theorem.

Example – 33

Find the sum of the divisors of 120. Generalise the result for an arbitrary natural number N.

Solution: The divisors of 120 are listed out below:

{1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60, 120}

The sum of these divisors is 360. We have to determine an elegant way to deduce this sumbecause we cannot repeat everytime the procedure of listing down all the factors and summingthem.For this purpose, we again resort to the use of the prime factorization form.

3 ' 1120 2 3 5= ⋅ ⋅The sum of the divisors will be

0 30 10 1

2 3 5i j k

ijk

S≤ ≤≤ ≤≤ ≤

= ⋅ ⋅∑

This notation is simply a shorthand which implies that we vary the integral indices i, j and k (in theirrespective allowed ranges) and this way we will have listed down all the factors and henceevaluated the required sum.



To generate the expression for the sum S, we can alternatively use the following method:

( )( )( )1 2 3 1 11 2 2 2 1 3 1 5S = + + + + +

Did you realize the trick? Writing S this way also gives rise to all the factors. You are urged toconvince yourself about this by expanding this expression and observing that all possible factorswill be generated.

Thus, S can now be simply evaluated as follows:

( ) ( ) ( )1 2 3 1 11 2 2 2 1 3 1 5S = + + + + +

15 4 6= × ×

360=

This is the same result that we got earlier!

To do the general case, assume that the prime factorization form of N is

1 21 2 ........ n

nN p p pαα α=

where the 'iα are all positive integers and the '

ip s are all primes. The sum Sf of all the factors willbe

( ) ( ) ( )1 22 2 21 1 1 2 2 21 ..... 1 ..... ......... 1 ..... n

f n n nS p p p p p p p p pαα α= + + + + + + + + + +

Example – 34

Find the sum of all the five-digit numbers that can be formed using the digits 1, 2, 3, 4 and 5 if no digit isrepeated.

Solution: This problem can be solved very easily if we view it from an individual digit’s perspective.Suppose that we only consider the digit “4”. How many numbers will there be with “4” in the unitsplace?

There are 24 numbers with 4 in the

units place because the remaining

four digits can be permuted among

the remaining 4 places in 4!=24 ways

4

4

4

''

From these 24 numbers, what is the total contribution of the digit “4” to the sum we are requiredto calculate? Since “4” is at the units place and it occurs 24 times, its contribution will be4 × 24 = 96



Similarly, there will be 24 numbers where “4” is at the tens place. The total contribution of “4”from these 24 numbers will be 4 × 240 = 960Proceeding in this way, we see that the total contribution of the digit “4” from all the 120 numbersthat can be possibly formed is:

( )4 24 240 2400 24000 240000+ + + +

( )4 24 1 10 100 1000 10000= × × + + + +

4 24 11111= × ×

This is the contribution to the sum from only the digit “4”. To calculate the entire sum S, wecalculate the contributions from all the five digits. Thus, the sum is

( )1 2 3 4 5 24 11111S = + + + + × ×

= 3999960

Example – 35 DEARRANGEMENTS

Find the number of ways in which 5 different letters can be taken out of their 5 addressed envelopes and putback into the envelopes in such a way that all letters are in the wrong envelope.

Solution: The problem of rearranging objects so that each object is assigned to a place different from itsoriginal place is referred to as the problem of dearrangments. Here, we need to find out thenumber of dearrangements possible with 5 letters and 5 envelopes.

Let us denote by Dn the number of dearrangements possible with n things. We will attempts ageneral solution, that is for an arbitrary n, and then substitute n = 5

Denote by Li the ith letter and by Ei, the original envelope of the ith letter.

Consider the envelope E1. It can be assigned a wrong letter in (n – 1) ways. Suppose that weassign the letter L2 to E1.

1 2

1 2

............

............

n

n

E E E

L L L(

The dearrangements that can now arise can be divided into two mutually exclusive classes:

(i) Those in which L1 is assigned to E2

(ii) Those in which L1 is not assigned to E2

If L1 is assigned to E2, we have the following configuration:

1 2

1 2

............

............

n

n

E E E

L L L(!

In this case, we have a remaining of (n – 2) letters which can be dearranged in Dn–2 ways.



Suppose the other case now, where we do not assign L1 to E2. In this case, we have (n – 1) lettersto dearrange (L1 will also be counted as a letter to be dearranged since it is being assigned to anenvelope other than E2). This can be done in Dn–1 ways.

Thus, if we give L2 to E1, the total dearrangements possible are 1 2.n nD D− −+

Since E1 can assigned a wrong letter in (n –1) ways, the overall total number of dearrangementsDn is

( )( )1 21n n nD n D D− −= − +

We have thus related the nth order 'dearrangements–coefficient' Dn to lower order coefficients.

We now have to somehow use this relation to obtain Dn only in terms of n. This is how we do it:

( )( )1 21n n nD n D D− −= − +

( ) ( )( )11 1 21 n

n n n nD nD D D−− − −⇒ − = − − ... (1)

But if we substitute ( )1n n→ − we get

( ) ( ) ( )( )1 2 2 31 1 2n n n nD n D D n D− − − −− − = − − − ... (2)

We substitute (2) back into (1) to get

( ) ( )( )21 2 31 2n n n nD nD D n D− − −− = − − −

( ) ( )( ) { }33 4 By the same process1 3n nD n D− −= − − −

'

( ) ( )22 11 2n D D−= − −

It is obvious that D1 = 0 since one letter cannot be dearranged while D2 = 1 because two letterscan be dearranged in only one way : by exchanging them.

Thus,

( ) 21 1 n

n nD nD −−− = − ... (3)

We still have not obtained a relation involving only Dn. We do it using (3) repeatedly

( ) ( )21 1 1n n

n nD nD −−− = − = −

( )( )1 1

! 1 ! !

nn nD D

n n n− −

⇒ − =− (Division by n!)



Now we substitute ( ) ( )1 , 2n n n n→ − → − .... successively and add the corresponding sidesof all the relations so obtained to get

( ) ( )

( )( )( )

( )1 2 21 1 1 1 1

......! 1! ! 1 ! 2 ! 2!

n n nnD D

n n n n

− −− − − −− = + + + +

− −

Since D1 = 0, we finally get

( )11 1 1! ........

2! 3! 4! !

n

nD nn

−= − + −

( )11 1 1! 1 ...

1! 2! 3! !

n

nn

−= − + − +

1The first two terms "1"and " "

1! are just added to make the

series look more sequenced

This is the number of de arrangements possible with n things For n = 5, we have

51 1 1 1 1D 5! 11! 2! 3! 4! 5!

= − + − + −

= 44

Thus, there are 44 ways to rearrange the letters back into their envelopes so that each letter goesto a wrong envelope.Since n = 5 is a small number, we could have worked out an alternative solution as follows:

Total No. of arrangements -No. of ways in which all letters go to correct envelopes - No. of ways in

No. of dearrangemets which one letter goes to incorrect envelope-No. of ways in which two letters

=

go to incorrect envelopes -.............................and so on

You are urged to work out the solution by this way yourself.

Example – 36

Prove that for 4,n ≥ the sum 1! 2! 3! ... !n+ + + + cannot be the square of a positive integer.

Solution: The approach lies in realizing the general properties of a perfect square. Note that a square willhave, at its unit place, a digit from amongst the set {1, 4, 5, 6, 9}. It cannot have a ‘3’ or an ‘8’ asits unit digit. Work it out yourself.



Now, for the given sum, the first four terms add to 33:

1! 2! 3! 4! 1 2 6 24+ + + = + + +

= 33

For any integer greater than 4, its factorial will always have a ‘0’ at its unit place. Can you seewhy?

This means that the given sum will have a ‘3’ at its unit place, so it cannot be a perfect square!

Example – 37

A set P contains x elements while Q is another set which contains y elements. How many functions :f P Q→exist which are (a) one - one (b) onto?

Solution: Obviously, the total number of functions possible is xy

(a) If the function is to be one-one, y, the number of elements in the co-domain of f, must begreater than or equal to x :

y x≥

In that case, the number of one-one functions will simply be ,yxP since you are creating a

mapping from x elements in P to x elements in Q, out of a total y elements in Q.

(b) For the function to be onto, the only constraint that needs to be satisfied is that each elementin the co-domain Q must have a pre-image, which means that x must be greater than or equalto y.

The remaining part is left to the reader as an exercise.

Example – 38

Four non-identical dice are rolled simultaneously. How many rolls are possible in which 6 shows up on at leastone dice?Solution: Its easiest to proceed by considering the complementary case:

Total number of rolls possible : 64 (Why ?)

Rolls in which no 6 shows up : 54 (Why ?)

Our answer is therefore 64 – 54.

Now, a very important issue which we leave to you to resolve: what if the four dice were identical?

Example – 39

In an examination, the maximum marks for each of three papers is n and that for the fourth paper is 2n. Findthe number of ways in which a candidate can get 3n marks.



Solution: We need to find the number of solutions to the non-negative integral solution

1 2 3 4 3 ,x x x x n+ + + = ... (1)

where 1 2 3, ,x x x all lie in [0, n], while 4x can vary upto 2 ,n i.e., 4 [0,2 ],x n∈ where all the 'ix s

are integers. Let us construct the polynomial ( )P x given by2 3 2( ) (1 ... ) (1 ... )n nP x x x x x x x= + + + + + + + +

Now, think hard ! The first bracket in ( )P x is actually three identical brackets 2(1 ... ).nx x x+ + + +When ( )P x is expanded , each of these three brackets can contribute at the most an nth powerin x, whereas the fourth bracket can contribute upto 2n in powers of x.

When ( )P x is expanded and the terms written out, note that 3nx will also be formed. But what

will be the coefficient of 3 ?nx It will be precisely the number of solutions to the equation (1) underconsideration !

Why ? Because, each time 3nx is formed is a unique combination of power contribution from eachof the four brackets. For example,

2

3

2

Take Take 1 Take 1 Take

Product formed =

( ) (1 ... )(1 ... )(1 ... )(1 ... )

nn

n

n n n n

xx

x

P x x x x x x x x x↓↓ ↓ ↓

= + + + + + + + + + +

"##########$##########%

In this example, we take the term 1( or x0) from the first two brackets, nx from the third and 2nxfrom the fourth. This is a unique combination in which 3nx can be generated, but this is only onesuch. There will be many more ways in which 3nx can be generated. How many ? Precisely, thecoefficient of 3nx when ( )P x is expanded !To find out the coefficient, you’ll have to wait until you are familiar with the Binomial / Multinomialexpansion.

Example – 40

In how many ways can one fill an m × n table with 1± such that the product of the entries in each row and eachcolumn equals –1?

Solution: Denote by ija the entries in the table. For 1 1i m≤ ≤ − and 1 1,j n≤ ≤ − we let 1ija = ± in an

arbitrary way. This can be done in ( 1)( 1)2 m n− − ways. The values for mja with 1 1,j n≤ ≤ − and for

ina with 1 1,i m≤ ≤ − are uniquely determined by the condition that the product of the entries in

each row and each column equals –1. The value of mna is also uniquely determined, but it isnecessary that

1 1

1 1

n m

mj inj i

a a− −

= =

= ⋅∏ ∏ (*)



If we denote by

1 1

1 1

m n

iji j

P a− −

= =

=∏∏

we observe that

1

1

1

( 1)n

nmj

jP a

−−

=

= −∏

and

1

1

1

( 1)m

min

i

P a−

−

=

= −∏

Hence (*) holds if and only if m and n have the same parity.



CONCLUSION : A MAGIC TRICK

For those of you interested in cards, here is a very interesting card trick that has its basis in an elementaryapplication of this chapter’s concepts.

A magician sends out his assistant into the audience with a well-shuffled standard deck of cards. The assistantasks 5 different people chosen at random from this audience to pick a card each from the deck. The magicianis on-stage and cannot see the cards drawn from the deck. The magician then asks the assistant to reveal4 of the 5 drawn cards while keeping the fifth card secret.

The magician then thinks for a while and tells the audience the number and suit on the fifth secret card ! Wehave to figure exactly how the assistant conveys the information about the fifth card to the magician. You canassume that no one is cheating!

The trick explainedSomehow, the assistant has been able to convey information about the fifth card to the magician, even thoughhe reveals only 4 of the 5 cards. The information must be embedded in the order in which the assistant revealsthe 4 cards. Thus, the magician and assistant must have pre-decided on a scheme whereby the magician candeduce the secret card from the order of the 4 revealed cards. Here is how this can be done.

Since 5 cards are drawn, at least 2 cards must be of the same suit. If there are more than 2 cards of the samesuit, the assistant simply chooses any 2 of them.

Now, one of these 2 cards will be revealed first by the assistant while the other card will become the secretcard. So, the magician knows from the suit of the first revealed card what the suit of the secret card is.

The first issue is that, of the 2 cards of the same suit, how does the assistant decide which one to reveal andwhich one to keep secret?

For that, the assistant refers to the following diagram (mentally of course!):

A2

3

4

5

6

789

10

J

Q

K

Fig - 17



In this circular arrangement, observe that the difference between any two members is at the most 6. Forexample, the difference between “7” and “Q” is 6 while that between “J” and “2” is 4

A2

3

4

5

6

789

10

J

Q

KA

23

4

5

6

789

10

J

Q

K

Fig - 18

4 step

s

6 ste

ps

Thus, of the 2 cards of the same suit, the assistant reveals that card from which it is possible to reach the othercard in at most 6 steps.

For example, if the assistant finds a “J” of Hearts and a “2” of Hearts in the 5 drawn cards, he will reveal the“J” of Hearts first and keep the “2” of Hearts secret since in our circular arrangement it is possible to reachfrom “J” to “2” in 4 steps.

Thus, till now the scheme is that the secret card is of the same suit as the first revealed card and lies within atthe most 6 steps from the first card, that is, the offset of the secret card from the first card is at the most 6.

We now require a scheme whereby the assistant can communicate the offset to the magician.

For this purpose, they pre-decide upon an order of all the 52 cards of the deck. For example, suppose theyorder the deck like follows:

A of 2 of 4 of A of 3 of 2 of 3 of A of A of K of clubs clubs clubs spadesclubs spades spades hearts diamonds diamonds

1 2 3 4 ... 14 15 16 ... 27 ... 40 ... 52

Increasing order

↓ ↓ ↓ ↓↓ ↓ ↓ ↓ ↓ ↓

→

Fig 19

Now, after the assistant has revealed the first card, he has 3 more cards to reveal. Those 3 cards can beordered according to the order of Fig 19. Thus, one of those 3 cards will be the ‘smallest’, one will be the‘middle’ card and one will be the ‘largest’. Label the 3 cards as S, M and L.



To communicate the offset to the magician, the assistant reveals the 3 cards in a particular order as shown inthe table below

Offset Order in which the assistant reveals the 3 cards1 S M L2 S L M3 M S L4 M L S5 L S M6 L M S

Since the maximum offset can be 6, the 6 permutations of S, M and L cover all the possible offsets!

Once the offset is communicated, the magician simply adds it to the first card to deduce the fifth, secret card!

Let us try this out with a particular example. Suppose that the following 5 cards are drawn: “A of spades” “2of Hearts” “3 of clubs” “7 of Hearts” “8 of clubs”.

The assistant picks 2 cards of the same suit. Here we have 2 options for the 2 cards. We can choose either.Suppose the assistant chooses “2 of Hearts” and “7 of Hearts”. The offset of “7” from “2” is 5 which is lessthan 6.

Therefore, “2 of Hearts” is the first card that the assistant reveals.

To communicate the offset, the assistant orders the three other cards (to be revealed) according to Fig. 19 asfollows:

"3 of clubs" "8 of clubs" "A of spades"S M L

< <

Since the offset required is 5, the assistant reveals these three cards (according to the Table on the previouspage) in the order L S M.

Thus, this is what the assistant says:

“2 of Hearts”

“A of spades”

“3 of clubs”

“8 of clubs”

The magician thinks for a while and says, “The fifth card is the 7 of Hearts !” And the audience gasps!



P & C [ ASSIGNMENT ]

[ OBJECTIVE ]

LEVEL - IQ. 1 The number of ways in which 4 balls can be selected from a bag containing 4 identical and 4 different

balls, is(a) 120 (b) 80 (c) 14 (d) 16

Q. 2 The maximum number of intersection points of n straight lines, none of which are parallel to each other,is

(a) 2n C (b) 2

n P (c) 2

2

n C(d) 2

n C n−

Q. 3 There are 4 bulbs in a room. The number of ways of illuminating the room, is(a) 16 (b) 15 (c) 24 (d) 4

Q. 4 There are n different books and each book has p copies. The number of selection of books from theseis

(a) ( )1 1np + − (b) ( )1 2np + − (c) ( )1 1n p + − (d) none of these

Q. 5 The value of 18 17 16 15 142 2 2 2 3C C C C C+ + + + is

(a) 173C (b) 18

3C (c) 1916C (d) none of these

Q. 6 The number of ways in which 6 men can be made to sit round a table in 6 numbered seats is(a) 5! (b) 6! (c) 6 6!× (d) none of these

Q. 7 The number of diagonals that can be formed by joining the vertices of an n-sided polygon, is

(a) 2n C n− (b) 3

nP (c) ( )1n n − (d) ( )2 1n C n− +Q. 8 The maximum number of intersection points of n circles, is

(a) 2n C (b) 2

n P (c) 2

2

n C(d) 2

n C n−

Q. 9 In an examination there are 5 multiple choice questions. Each question has four choices of which onlyone is correct. The number of ways in which an examinee can get at least one answer wrong, is(a) 54 1− (b) 45 1− (c) 5!4! 1− (d) 5

4 1C −

Q. 10 The exponent of 4 in 200!, is(a) 86 (b) 101 (c) 47 (d) 65

Q. 11 The number of different messages by signals with three dots and two dashes is(a) 120 (b) 12 (c) 10 (d) 20

Q. 12 If 1 36, 84n nr rC C− = = , 1 126n

rC + = then the value of r is

(a) 1 (b) 2 (c) 3 (d) none of these



Q. 13 Five persons including one lady are to deliver lectures to an audience. The organiser can arrange thepresentation of their lectures, so that the lady is always in the middle, in(a) 5

5P ways (b) 444. P ways (c) 4! ways (d) 5

4C ways

Q. 14 The number of ways in which a set A where n(A) = 12 can be partitioned in three subsets P, Q, R eachof 4 elements if P Q R A∪ ∪ = , ,P Q φ∩ = ,Q R φ∩ = R P φ∩ = , is

(a) ( )312!4! (b) ( )4

12!3! (c) ( )3

12! 1.3!4! (d) none of these

Q. 15 In a test of 10 multiple-choice questions of one correct answer, each having 4 alternative answers, thenumber of ways to put ticks at random for the answers to all the questions is(a) 104 (b) 410 (c) 104 4− (d) 410 10−

Q. 16 The number of positive integral solutions of the equation 50,x y z+ + = is

(a) 512C (b) 50

3C (c) 492C (d) 48

3C

Q. 17 The number of zeros at the end of 50!, is(a) 17 (b) 5 (c) 12 (d) 47

Q. 18 The number of ways in which 9 different objects can be distributed to three different people giving 2,2 and 5 objects to each of them, is(a) 328 (b) 984 (c) 756 (d) 2268

Q. 19 If m parallel straight lines intersect n parallel straight lines, then the number of parallelograms thusformed, is(a) nm (b) 2

m n C+ (c) mn (d) 2 2m nC C×

Q. 20 There are 2 points on a line, 3 points on another line and 4 points on yet another line. The total numberof triangles that can be formed by joining these points, is(a) 30 (b) 205 (c) 79 (d) 85

[ LEVEL - II ]

Q. 1 The number of divisors of 1008 of the form 4 2( 0,n n n+ ≥ ∈ N) , is(a) 5 (b) 6 (c) 11 (d) 7

Q. 2 Numbers are formed using the digits 1, 1, 2, 2, 2, 2, 3, 4, 4 all at a time. The number of such numbersin which the odd digits occupy even places, is

(a) 162 (b) 175 (c) 180 (d) 160

Q. 3 The number of ways of distributing 10 different toys among 4 children C1, C2, C3, C4 such that C1 andC2 get 2 toys each and C3 and C4 get 3 toys each, is

(a) ( ) ( )2 210!

2! 3! (b) ( ) ( ) ( )2 2 210! 4!

2! 3! 2!× (c) ( ) ( )2 210 10

2 3C C (d) none of these



Q. 4 There are n different colours and p balls of each colours. The number of ways in which one or morethan one ball can be chosen, is(a) ( )1 1pn + − (b) ( )1 1np + − (c) 1p

nC − (d) 1 1pnC+ −

Q. 5 A four digit number is formed using the digits 1, 2, 3, 4, 5. The total number of numbers which have atleast one digit repeated, is(a) 625 (b) 500 (c) 120 (d) 505

Q. 6 The number of integral solutions of the equation ( )1 2 3 1 2 30 , , 3x x x x x x+ + = ≥ − , is

(a) ( )33! (b) ( )39!3! (c) 11

3C (d) 112C

Q. 7 The number of natural numbers less than 2000 that can be formed using the digits 0, 1, 2, 3, 4 withoutrepeating any digit, is

(a) 78 (b) 102 (c) 88 (d) 92

Q. 8 In the identify ( )( ) ( )0

!1 2 ...

nk

k

A nx k x x x x n=

=+ + + +∑ the value of kA is

(a) nkC (b) 1

nkC + (c) ( )1 .k n

kC− (d) ( ) 111 .k n

kC−−−

Q. 9 The number of times the digit 1 will occur in natural numbers less than 1000, is

(a) 250 (b) 300 (c) 350 (d) 450

Q. 10 The number of ways in which m prizes can be distributed among n players if each player can receiveany number of prizes, is

(a) nm (b) mn (c) mnC (d) n

mc

Q. 11 There are n identical seats (unmarked) around a circular table. The number of ways in which ( )m m n<people can be seated, is

(a) ( )1 !m − (b) ( )( )

1 !!

nn m

−− (c) ( )1 !n m− + (d) ( )1 !n

mC m −

Q. 12 The number of non-negative integral solutions of the equation 3 10x y z+ + = , is

(a) 26 (b) 10 92 3C C× (c) 160 (d)

10!2!3!

Q. 13 There are p points in space, no four of which are in the same plane with the exception of q points whichare all in the same plane. The number of different planes determined by the points is

(a) ( )( ) ( )( )1 2 1 2

6 6p p p q q q− − − −

− (b) ( )( ) ( )( )1 2 1 21

6 6p p p q q q− − − −

− +

(c) ( )( )( ) ( )( )( )1 2 3 1 2 3

124 24

p p p p q q q q− − − − − −− + (d) none of these



Q. 14 The number of ways in which 8 candidates 1 2 3 8, , ,.....,A A A A can be ranked such that 3A is always

comes before 4A and 4A always comes before 5A , is

(a) 862 P× (b)

8!6 (c) 8! (d)

8!3

Q. 15 The number of ways in which 8 candidates 1 2 8, ,.....,A A A can be ranked such that 4A is always

above 5A , is

(a) 862 P× (b) 8

6P (c) 8! (d) 8!3

Q. 16 The number of triangles that can be formed from n points of which m are collinear, is(a) ( )3

n C m n− + (b) 3n C (c) 3 3

n mC C− (d) 3n m C−

Q. 17 The number of ways in which a sum of 10 can be obtained by throwing a dice thrice, is(a) 21 (b) 27 (c) 6 (d) 15

Q. 18 There are 6 identical blue balls and 6 identical red balls. The number of ways in which 6 balls out of thegiven 10 balls can be arranged in a row such that the number of blue balls is equal to the number of redballs, is(a) 6 (b) 64 (c) 20 (d) 9

Q. 19 The sum of the digits in the units place of all numbers formed with the digits 1, 1, 2, 3 when taken all ata time, is(a) 21 (b) 17 (c) 32 (d) 28

Q. 20 If n objects are arranged in a row, then the number of ways of selecting three of these objects so thatno two adjacent object is selected, is

(a) 3 2n nC C− (b) ( )3

!3!n

(c) 3n C (d) 2

3n C−



[ SUBJECTIVE ]

Q. 1 A letter lock consists of 4 rings each marked with 15 different letters. In how many ways is it possibleto make an unsuccessful attempt to open the lock?

Q. 2 Show that the greatest value of 2nrC for 0 2≤ ≤r n is 2n

nC

Q. 3 There are an unlimited number of identical balls of four different colors. How many arrangements of atthe most 8 balls in a row can be made by using them?

Q. 4 How many different numbers greater than 5000 can be formed with the digits 1, 2, 5, 9, 0 whenrepetition of digits is not allowed?

Q. 5 A polygon has 44 diagonals. How may sides does it have?

Q. 6 There are m points on one straight line AB and n points on another straight line AC, none of them beingA. How many triangles can be formed with these points as vertices (excluding A)?

Q. 7 In how many ways can five different rings be worn in four fingers with at least one ring in each finger?

Q. 8 m men and n women are to be seated in a row so that no two women sit together. If m > n, show that

the number of ways in which they can be seated is ( )

( )! 1 !

1 !+

− +m mm n

.

Q. 9 How many four digit numbers with distinct digits can be formed using the digits 0, 1, 2, 3, 4, 5 whichare(a) divisible by 3? (b) divisible by 6?

Q. 10 How many words can be formed by taking 4 letters at a time out of the letters of the wordMATHEMATICS?

Q. 11 Find the total number of ways of selecting 5 letters from the letters of the word INDEPENDENT.

Q. 12 Two numbers a and b are chosen from the set {1, 2, 3, ....., 3n}. In how many ways can these integersbe selected such that

(a) a2 – b2 is divisible by 3? (b) a3 + b3 is divisible by 3?

Q. 13 What is the number of non-negative integral solutions of the equation 1 2 3 44 20?+ + + =x x x x

Q. 14 An exam consists of four papers. Each paper has a maximum of m marks. Show that the number of

ways in which a student can get 2m marks in the exam is ( )21 2 4 33+ + +m m m



Q. 15 Find the total number of positive unequal integral solutions of the equation 20+ + + =x y z w

Q. 16 In how many ways can 10 persons take seats in a row of 24 fixed seats so that no two persons takeconsecutive seats?

Q. 17 There are 15 seats in a row numbered 1 to 15. In how many ways can 4 persons sit in such a way thatseat number 6 is always occupied and no two persons sit in adjacent seats?

Q. 18 In how many ways can 2n + 1 identical balls be placed in distinct boxes so that any two boxestogether will contain more balls than the third?

Q. 19 12 seats are to be occupied by 4 people. Find the number of possible arrangements if

(i) no two persons sit side by side.

(ii) there should be at least two empty seats between any two persons.

(iii) each person has exactly one neighbour.

Q. 20 The sides of a triangle are a, b, c inches where a, b, c are integers and ≤ ≤a b c . If c is given, show

that the number of different triangles that can be formed is ( 2)4+c c if c is even and

2( 1)4+c

if c

is odd.



[ TRY YOURSELF - I ]

1. 161280 2. 120 3. 10 4. 120960 5. 3606. 180 7. 1200 8. 246 9. 26 10. 35

SUBJECTIVE

1. 154 – 1 3. 87380 4. 48 5. 11 6. ( )22

mn m n+ −

7. 480 9. 96, 52 10. 2454 11. 72 12. 2 25 3 3,2 2

n n n n− −

13. 536 15. 552 16. 1510P 17. 3072 18. ( )1

2n n +

19. 9 6 94 4 2, , 4!P P C×

[ ASSIGNMENT ]

LEVEL - II

1. b 2. c

3. a 4. b

5. d 6. d

7. d 8. c

9. b 10. b

11. b 12. a

13. b 14. b

15. d 16. c

17. b 18. c

19. a 20. d

OBJECTIVE

LEVEL - I

1. d 2. a

3. b 4. a

5. c 6. b

7. a 8. b

9. a 10. d

11. c 12. c

13. c 14. c

15. a 16. c

17. c 18. d

19. d 20. c

P & C / ANSWER



APPENDIX : BINOMIAL THEOREM

Binomial theorem is something that has been known to mathematicians since many centuries ago. In thisintroduction, we’ll trace the origins of this theorem to the coefficients we obtain when we expand any binomialterm raised to an integral power.Consider the following expansions, which can be verified by direct multiplication:

( )0 1x y+ =

( )1x y x y+ = +

( )2 2 22x y x xy y+ = + +

( )3 3 2 2 33 3x y x x y xy y+ = + + +

( )4 4 3 2 2 3 44 6 4x y x x y x y xy y+ = + + + +

( )5 5 4 3 2 2 3 4 55 10 10 5x y x x y x y x y xy y+ = + + + + +

( )6 6 5 4 2 3 3 2 4 5 66 15 20 15 6x y x x y x y x y x y xy y+ = + + + + + +

' and so on

Do you notice anything special about these expansions, in particular, any general rule or trend these expansionsfollow that might enable us to expand ( )nx y+ directly for a general n ? First of all, notice that the number of

terms in each expansion is one more than the power of the binomial term. For example, ( )5x y+ has 6 terms.

However, mathematicians long back also realized another important fact, namely, the relation between thecoefficients obtained upon expansion. To see what this relation is, let us write the coefficients in the following‘triangular’ pattern:

1

1

1

1

1

1

1

1

1

1

1

1

12

3 3

4 4

5 51010

156 20

6

15 6

Fig - 20

n = 0

n = 1

n = 2

n = 3

n = 4

n = 5

n = 6



Do you observe any relation between the various coefficients. If not, consider this same arrangement in aslightly modified form:

1

1

1

1

1

1

1

1

1

1

1(1+1)

(1+2) (2+1)

(1+3) (3+1)

(1+4) (4+1)(6+4)(4+6)

(3+3)

Fig - 21The ‘rule’ for constructing this triangular pattern should be pretty obvious now. All edge-numbers are 1. Anyother is obtained by adding the number directly above and to the left with the number directly above and to theright, as in Fig-34 Extending this process gives us all the ‘binomial coefficients.’ This geometrical arrangementof the binomial coefficients in a triangle is called Pascal’s triangle. The figure below shows a Pascal trianglecontaining the coefficients upto n = 15.

11

11

11

11

11

11

11

11

11

11

11

11

11

1

23 3

4 6 45 10 10 5

6 15 20 15 67 21 35 35 21 7

8 28 56 70 56 28 89 26 84 126 126 84 26 9

10 45 120 210 252 210 120 45 1011 55 165 330 462 462 330 165 55 11

12 66 220 495 792 924 792 495 220 66 1213 78 286 715 1287 1716 715 2861716 1287 78 13

1 1414 91 364 1001 2002 3003 3003 10013432 2002 364 91 1

Pascal's triangle,upto = 15n

Fig - 22

If we denote the (i + 1)th number at the nth level by Tn, i , then we have, 1 1, 1, 1n i n i n iT T T+ − − += + ... (1)

Later on, when we actually write , 1n iT + in terms of combinational notation (in fact, we’ll see that , 1n iT +

corresponds to niC ), we will immediately understand that (1) is equivalent to

1 11

n n ni i iC C C− −

−= +which, as we already know from the last chapter on P & C, is true.



Section - 1 Section - 1BINOMIAL THEOREM, POSITIVE INTEGRAL INDEX

Let us now consider more formally the binomial theorem. We need to expand ( )nx y+ , where x, y are twoarbitrary quantities, but n is a positive integer.

In particular, if we write

( ) ( )( )( ) ( ) ( )... timesnx y x y x y x y x y n+ = + + + + ... (1)

we need to find out the coefficient of i jx y . Note that i j+ must always equal n, so that we can write ageneral term of the expansion (without the coefficient) as r n rx y − so that 0 r n≤ ≤ .

Now, to find the coefficient of r n rx y − , note that we need the quantity x, r times, while y is needed n – r times.Thus, in (1), r n rx y − will be formed whenever x is ‘contributed’ by r of the binomial terms, while y is ‘contributed’

by the remaining n - r of the binomial terms. For example, in the expansion of ( )5x y+ , to form 2 3x y , weneed x from 2 terms and y from 3:

( ) ( )( )( )( )( )5x y x y x y x y x y x y+ = + + + + +

take x

take y

One of the ways to form x y2 3

Fig - 23

How many ways are there to form 2 3x y ? In other words, how many times will 2 3x y be formed? The number

of times 2 3x y is formed is what is the coefficient of 2 3x y . That number, which would be immediately obvious

to the alert reader, is simply 52C . Why? Because this is the number of ways in which we can select any 2

binomial terms from 5. These 2 terms will contribute x. The remaining will automatically contribute y.

In the general case of ( )nx y+ , we see that the coefficient of r n rx y − would be nrC . (which is infact the same

as nn rC − ). Thus, the general binomial expansion is

( ) 1 2 20 1 2 ...n n n n n n n n n

nx y C x C x y C x y C y- -+ = + + + +

( )0

nn n n r r

rr

x y C x y−

=

⇒ + = ∑



The coefficients niC are called the binomial coefficients, for a reason that should now be obvious.

Note that the ( )th1i + coefficient in this expansion is niC , which now explains the relation

, 1 1, 1, 1n i n i n iT T T+ − − += +

we observed in the Pascal triangle; this relation simply corresponds to

1 11

n n ni i iC C C− −

−= +

Also, the binomial coefficients of terms equidistant from the beginning and the end are equal, because we haven n

r n rC C −= . The general term of expansion, n n r rrC x y− , is the (r + 1)th term from the beginning of the expansion

and is conventionally denoted by 1rT + , i.e.

1n n r r

r rT C x y−+ =

Since we have (n + 1) terms in the general expansion, we see that if n is even, there will be an odd number ofterms, and thus there will be only one middle term, which would be / 2 / 2

/ 2n n n

nC x y . For example,

( )4 4 3 2 2 3 4

onlyonemiddle term

4 6 4x y x x y x y xy y+ = + + + +

On the other hand, if n is odd , then there will be an even number of terms in the expansion, and thus there will

be two middle terms, namely 1 1 1 1

2 2 2 21 1

2 2

andn n n n

n nn nC x y C x y

+ − − +

− −

For example;

( )5 5 4 3 2 2 3 4 5

Two middle terms5 10 10 5x y x x y x y x y xy y+ = + + + + +

Example – 1

Find the middle term(s) in the expansion of 91x

x −

Solution: Since n = 9, there will be 10 terms in the expansion, which means that there will be 2 middleterms in the expansion, the 5th and the 6th:

( )4

9 495 4

1 126T C x xx

− − = =

( )9 596 5

5

1 126T C xx x

− − − = =



Example – 2

Is there any term in the expansion of 1022 1

5x

x

−

that will be independent of x?

Solution: The general term in the expansion is

102

101

2 15

r r

r rxT C

x

−

+ − =

( )10

20 210 22 15

r rrrrC x

−− − = −

( )10 52010 22 1

5

r rr

rC x−

− = −

Thus, for the term that is independent of x, we have

520 02r− =

8r⇒ =

Thus, the term free of x is the 9th term given by

( )2

8109 8

2 3615 5

T C = − =

Example – 3

Evaluate the sum 0 1 2 ...n n n nnC C C C+ + + + .

Solution: We have already evaluated this sum in the chapter on P & C. That approach was as follows: thissum basically counts the number of all sub-groups of a set of size n; this can also be counted byfocusing on each element of the set, which has two corresponding choices - you either include itinto your sub-group or you don’t, which means that the total number of ways to form sub-groupsis 2 × 2 × 2 .... n times = 2n. The sum of the binomial coefficients therefore equals 2n.

Here, we evaluate the same sum using a binomial approach. Consider the following expansion:

( ) 2 30 1 2 31 ...n n n n n n n

nx C C x C x C x C x+ = + + + + +

If we put x = 1, we simply obtain

0 1 22 ...n n n n nnC C C C= + + +



Thus, the same result is obtainable from both a combinatorial and a binomial approach.

We can also derive another useful result by putting x = –1 in the above relation, so that weobtain

( )0 1 2 30 ... 1 nn n n n nnC C C C C= − + − + + −

0 2 4 1 3 5... ...n n n n n nC C C C C C⇒ + + + = + + +

This states the sum of the even-numbered coefficients is equal to the sum of the odd-numberedcoefficients. Can you prove this using a combinatorial approach?

As an exercise, prove the following relations:

1 20 1 22 2 2 ... 3n n n n n n n n

nC C C C− −+ + + =

( ) ( )20 1 2 ... 1 1n nn n n n n

nC C x C x C x x− + − + − = −

( )31 20 2 3

1 1...2 2 2 2 2

n nnn nnn

n n

CCC CC−

− + − + + =

Example – 4

(a) What is the greatest coefficient in the expansion of ( )nx y+ ?

(b) What is the greatest term in the expression of ( )nx y+ ?

Solution: (a) For this part, we basically need to only determine ( )max for 0 ;nrC r n≤ ≤ x and y have

no role to play in this part.To find the greatest coefficient, consider the following ratio:

( ) ( )

( )

1

!1 ! 1 !

!! !

nr

nr

nr n rCq nC

r n r

+ + − −= =

−

1n rr

−=+

Thus,

1 11

n rqr

−> ⇒ >+

1n r r⇒ − > +

12

nr −⇒ <



Similarly,

1 11

n rqr

−< ⇒ <+

1n r r⇒ − < +

1

2nr −⇒ >

Thus,

1

1

1whenever2

1and whenever2

n nr r

n nr r

nC C r

nC C r

+

+

− > < − < >

... (1)

If n is odd, we have

1 32 2

n nn nC C− −>

and 3 12 2

n nn nC C+ +<

Also, since

1 12 2

n nn nC C− +=

we see that for odd n, the two middle coefficients are the greatest. This can be verified byconsidering the following expansion:

( )5 5 4 3 2 2 3 4 5

The two middle coefficients are the greatest for odd

5 10 10 5

n

x y x x y x y x y xy y+ = + + + + +!(

If n is even, (1) gives

12 2

n nn nC C

−>

and 12 2

n nn nC C

+<

In this case therefore, the greatest coefficient is the single middle coefficient 2

nnC . Lets verify

this again:

( )6 6 5 4 2 3 3 2 4 5 6

The single middle coefficient is the greatest for even

6 15 20 15 6

n

x y x x y x y x y x y xy y+ = + + + + + +!



(b) To find the greatest term, we must also consider x and y. We again follow the approach ofpart (a):

11 1

1

n n r rr r

n n r rr r

T C x yqT C x y

−+

− + −−

= =

( )1

.n r y

r x− +

=

Observe that

( )11 1

n r yqr x

− +> ⇒ ⋅ >

( )10

n y x yrx y r x+ +⇒ − ⋅ > +

If ( )1n y

x y++ is an integer m , which must lie in (0, n ], we see that there are two greatest terms

Tm and Tm + 1. (Why). Here’s the explanation:

We have 1q > for 1 r m≤ < and 1q < for r m>

1r rT T +⇒ < for 1 r m≤ < and 1r rT T +> for 1m r n+ ≤ ≤

1 1 2 1, ,m m m m m mT T T T T T− + + +⇒ < > =

1andm mT T +⇒ are the two greatest terms

Now, if ( )1n y

x y++ is a non-integer, assume

( )1n ym

x y+

= + .

We now have

1q > for 1 r m≤ ≤ and 1q < for r m>

1r rT T +⇒ < for 1 r m≤ ≤ and 1r rT T +> for r m>

1 1 2,m m m mT T T T+ + +⇒ < >

1mT +⇒ is the greatest term.



Example – 5

How will you expand the multinomial expression ( )1 2 ... nmx x x+ + + ?

Solution: We will approach this problem using combinatorics. Note that a general term of the expansionwould be of the form (without the coefficient)

31 21 2 3 .... mn nn n

mx x x x ... (1)

where the various powers must always sum to n (why?).

i.e.,

1 2 3 ... mn n n n n+ + + + =

Now, to evaluate the coefficient of the term in (1), we consider the multinomial expression inexpanded form:

( )( ) ( )1 2 1 2 1 2

times

... ... .............. ...m m m

n

x x x x x x x x x+ + + + + + + +"###########$###########%

To generate the term in (1), we must get x1 from n1 terms, x2 from n2 terms and so on. Let us findthe number of ways in which this can be done.

First select those n1 multinomials that will contribute x1 : this can be done in 1

nnC ways. Now,

from the remaining ( )1n n− multinomials, select those n2 multinomials that will contribute x2 : thiscan be done in ( )1

2

n nnC− ways. Continuing this process, we see that the number of ways to get

1 1 2from ,x n x from n2 ... and so on, that is, the number of times the term in (1) will be generatedin the expansion is

( ) ( )1 1 2

1 2 3...n n n n nn

n n nC C C− − −× × ×

( )( )( )

( )( )

1 1 2

1 1 2 1 2 3 1 2 3

! !! ...! ! ! ! ! !

n n n n nnn n n n n n n n n n n n

− − −= × × ×

− − − − − −

1 2

!! !... !m

nn n n

=

This is what is known as the general multinomial coefficient. The multinomial expansion cannow be written compactly as

( ) 1 21 2 1 2

1 2

!... ...! !... !

mn nn n

m mm

nx x x x x xn n n

+ + + =∑



where the summation is carried out over all possible combinations of the ni's such that in n=∑ .

For example, in ( )41 2 3x x x+ + , let us consider some terms in the expansion:

Multinomial Term Coefficient

( + + )x x x1 2 34

21 2 3x x x

31 2x x

42x

1 2 3x x x2

4!2!1!1!

= 12

4!3!1!

= 4

4!4!

= 1

4!1!1!2!

= 12etc

Example – 6

Find the coefficient of x4 in the expansion of 10

2

21 xx

+ − .

Solution: From the previous example, the general term in the expansion will be

3

212

1 2 3

10! 2(1) ( )! ! !

nnn x

n n n x−

= 2 3 32

1 2 3

10 ( 2)! ! !

n n nxn n n

− −

where 1 2 3n n n+ + must be 10.

Now, x4 is generated whenever 2 32 4.n n− = The possible values of the triplet 1 2 3( , , )n n n cannow simply be listed out:

1 2 3( , , ) (6, 4, 0), (3, 6,1), (0, 8, 2)n n n ≡

Thus, the (total) coefficient of x4 is

0 1 210! 10! 10!( 2) ( 2) ( 2)6!4!0! 3!6!1! 0!8! 2!

− + − + −

= – 1290 (verify)



Section - 2 Section - 1DIFFERENTIATION & INTEGRATION TECHNIQUES

The techniques of calculus enable us to sum a lot of series involving binomial coefficients. This is the subject ofthis section.Suppose that we have to evaluate the sum S given by

1 2 32 3 ......n n n nnS C C C n C= + + + +

From now on, to avoid clutter, we’ll write nrC as simply Cr, where the upper index n should be understood

to be present. Thus,

1 22 ..... nnS C C C= + + +

rrC=∑This series can be generated using a manipulation involving differentiation, as follows:Consider the binomial expansion

20 1 2(1 ) ......n n

nx C C x C x C x+ = + + + +

If we differentiate both sides with respect to x, look at what we’ll obtain:1 2 1

1 2 3(1 ) 2 3 .....n nnn x C C x C x nC x− −+ = + + + +

Now, all that remains is to substitute x = 1, upon which we obtain:1

1 2 32 2 3 .....nnn C C C nC−⋅ = + + + +

This is what we were looking for. Thus, 12nS n −= ⋅

Had we substituted 1x = − , we would’ve obtained1

1 2 30 2 3 ....... ( 1)nnC C C nC−= − + + −

Thus, we have evaluated another interesting sum.Suppose that we now wish to evaluate S1 given by

1 21 0 ....

2 3 1nCC CS C

n= + + + +

+The alert reader would immediately realize that integration needs to be applied here. How exactly to do so isnow described. Consider again the expansion.

20 1 2(1 ) .....n n

nx C C x C x C x+ = + + + +

If we integrate this with respect to x, between some limits say a to b, we obtain

1 2 3 1

0 1 2(1 ) ....

1 2 3 1

b b b bn nb

naa a a a

x x x xC x C C Cn n

+ ++ = + + + ++ +



To generate the sum 1,S a little thought will show that we need to use a = 0, b = 1, so that we obtain

11 2

02 1 .....

1 2 3 1

nnCC CC

n n

+ − = + + + ++ +

Thus, S1 equals 12 1

1

n

n

+ −+

Try some other values for a and b and hence generate other series on your own. Be as varied as you can inchoosing these limits.

Example – 7

Find the sum S given by

2 2 2 21 2 31 2 3 .... nS C C C n C= ⋅ + ⋅ + ⋅ + + ⋅

Solution: We have to plan an approach wherein we are able to generate r2 with Cr. We can generate one rwith every Cr, as we did earlier, and which is now repeated here:

20 1 2(1 ) ......n n

nx C C x C x C x+ = + + + +

Differentiating both sides with respect to x, we have

1 2 11 2 3(1 ) 2 3 ......n n

nn x C C x C x nC x− −+ = + + + +

Now we have reached the stage where we have an r with every Cr. We need to think how to getthe other r. If we differentiate once again, we’ll have r(r – 1) with every Cr instead of r2(understandthis point carefully). To ‘make-up’ for the power that falls one short of the required value, wesimply multiply by x on both sides of the relation above to obtain:

1 2 31 2 3(1 ) 2 3 ....n n

nnx x C x C x C x nC x−+ = + + + +

It should be evident now that the next step is differentiation:

2 1 2 2 2 2 11 2 3( 1) (1 ) (1 ) 2 3 ....n n n

nn n x x n x C C x C x n C x− − −− + + + = + ⋅ + ⋅ + + ⋅

Now we simply substitute x = 1 to obtain2 1 2 2 2

1 2 3( 1) 2 2 2 3 .....n nnn n n C C C n C− −− ⋅ + ⋅ = + ⋅ + ⋅ + + ⋅

The required sum S is thus2 1( 1) 2 2n nS n n n− −= − ⋅ + ⋅

{ }22 ( 1) 2nn n−= ⋅ − +

2( 1) 2nn n −= + ⋅



Example – 8

Evaluate the following sums:

(a) 0 2 41 .......

1 3 5C C CS = + + + (b) 3 51

2 .......2 4 6

C CCS = + + +

Solution: The first sum contains only the even-numbered binomial coefficients, while the second containsonly odd-numbered ones. Recall that we have already evaluated the sum S given by

11 2

02 1......

2 3 1 1

nnCC CS C

n n

+ −= + + + + =+ +

Note that S is the sum of S1 and S2, i.e.,1

1 22 1

1

n

S Sn

+ −+ =+

Thus, if we determine S1, S2 is automatically determined, and vice-versa. Let us try to determine S1first.(a) Consider again the general expansion

20 1 2(1 ) ....n n

nx C C x C x C x+ = + + + +

Integrating with respect to x, we have (we have not yet decided the limits)

1 2 3 1

0 1 2(1 ) .....

1 2 3 1

b b b bn nb

naa a a a

x x x xC x C C Cn n

+ ++ = + + + ++ +

Since we are trying to determine S1 which contains only the even-numbered terms, we have tochoose the limits of integration such that the odd-numbered terms vanish. This is easily achievable

by setting a = – 1 and b = 1 (understand this carefully). Thus, we have1

2 40

2 2 ....1 3 5

n C CCn

+ = + + + + which implies that

12

1

n

Sn

=+

(b) S2 is now simply given by

2 1S S S= −

12 1 2

1 1

n n

n n

+ −= −+ +

2 1

1

n

n−=+



Section - 3 Section - 1MISCELLANEOUS TECHNIQUES

Not all questions can be subjected to the method(s) described earlier. For example, consider the sum S givenby

0 1 1 2 2 3 1...... n nS C C C C C C C C−= + + + +

Let us first go through a combinatorial approach, using the observation that 0 1 1,n nC C C C −= = and so on, sothat S can be rewritten as

1 1 2 2 3 1......n n n nS C C C C C C C C− −= + + + +

Consider a general term of this sum, which is of the form 1n r rC C− + . We can think of this as the number of waysof selecting (n – r) boys from a group of n boys and (r + 1) girls from a group of n girls. The total number ofpeople we are thus selecting is ( ) ( 1) ( 1)n r r n− + + = + . Therefore, S represents the total number of ways of

selecting (n + 1) people out of a group of 2n, so that S is simply 21

nnC + .

Now to a binomial approach. This will involve generating the general term 1r rC C + somehow, which is the

same as 1n r rC C− + . Consider the general expansion of (1 )nx+ .

20 1 2(1 ) .....n n

nx C C x C x C x+ = + + + + ...(1)

We have to have the terms 1 1 2,n nC C C C− and so on, which suggests that we write (1) twice, but in the secondexpansion we reverse the terms, multiply, and see what terms contain the (combinations of) coefficients werequire.

(1 ) + x = C + C x + C x + ..........+ C xn n0 1 2

2n

(1 ) + x = C x + C x + C x + .....+ C x + Cn n n nn n n – 1 – 2 1 0

–1 – 2

Multiplying, we find on the left hand side we have 2(1 ) nx+ , while on the right hand side, the terms containing

the (combinations of) coefficients we want will always be of the form 1( ) nx + , that is, the power of x will be

(n + 1). No other terms will contain 1nx + , verify this for yourself. Thus, the sum 1 1 2 1........n n nC C C C C C−+ + +

is actually the total coefficient of 1nx + on the right hand side, and from the left hand side we know that thecoefficient of 1nx + would be simply 2

1n

nC + . Thus, 21

nnS C +=

A very similar approach could have been:

(1 + ) = + C + + ........ + x C x C x C xn n0 1 2

2n

1 + 1x

n

= C C0 1+ 1x + C21x2 + ........ + Cn

1xn



Thus, S = Coefficient of x in 1(1 ) 1

nnx

x + +

= Coefficient of x in 2(1 ) n

nx

x+

= Coefficient of 1nx + in 2 21(1 ) n n

nx C ++ =

Example – 9

Find the sum S given by2 2 2 20 1 2 ...... nS C C C C= + + + +

Solution: Note that S can be rewritten as

0 1 1 2 2 0......n n n nS C C C C C C C C− −= + + + +

Using a combinatorial approach, the sum should be immediately obvious to the alert reader as2n

nC . In brief, this is because the right hand side represents, as an example, the total number ofways of selecting n people from a group of n boys and n girls, etc.

Now, we discuss the binomial expansion approach:

2 30 1 2 3

1 21 2 0

(1 ) .....

(1 ) .....

n nn

n n n nn n n

x C C x C x C x C x

x C x C x C x C− −− −

+ = + + + + +

+ = + + + +)* * *

Thus, we observe that the required sum is the coefficient of xn in 2(1 ) nx+ , which is simply 2nnC .

Example – 10

Find the sum S given by

1 20 1 2 ........n n n n r

rS C C C C+ + += + + + +

Solution: We have already evaluated this sum in P & C; here we’ll use a binomial approach. Note that

Coeff. of in (1 )n r n n rrC x x+ += +

⇒ ( )Coeff. of in (1 )n r n n rrC x x+ +∑ = ∑ +

= Coeff. of in (1 )n n rx x +∑ +



Thus,

1Coeff. of in (1 ) (1 ) ...... (1 )n n n n rS x x x x+ + = + + + + + +

{ 2Coeff. of in (1 ) 1 (1 ) (1 ) ...... (1 )n n rx x x x x = + + + + + + + +

{ }1(1 ) (1 ) 1

Coeff. of inn r

nx x

xx

++ + −=

{ }1 1Coeff. of in (1 ) (1 )n n r nx x x+ + += + − +

1n rrC+ +=

⇒ 1n rrS C+ += , which is the same result we obtained in P & C.



TRY YOURSELF - I

Q 1. Find the sum of all the rational terms in the expansion of ( )121/ 4 1/33 4+ .

Q. 2 Prove that three consecutive terms in a binomial expansion can never be in G.P.

Q. 3* (a) Show that the integral part of ( )675 5 11+ is even

(b) Show that the integral part of ( )8 3 7n

+ is odd

Q. 4 Use the binomial theorem to show that 7103 when divided by 5 leaves a remainder 3.Q. 5 Find the coefficient of x301 in the expansion of

500 499 2 498 500(1 ) (1 ) (1 ) ......x x x x x x+ + + + + + +

Q. 6* Suppose 2(1 )nx x+ + is written in expanded form, i.e., in the form 2

0

nr

rr

a x=∑ . Show that

2r n ra a −=

Q. 7* Use the binomial theorem to show that 323232 when divided by 7 leaves the remainder 4.Q. 8 Prove the following relations:

(a) 2 2 1 2 20 1 2 ....... ( 1) 1n n n n n n n n n

n n n n nC C C C C C C C− −⋅ − ⋅ + ⋅ − + − ⋅ =

(b) 2 2 2 2 40 1 2 ..... 2n n n n n n n

n n nC C C C C C− −⋅ − ⋅ + ⋅ − =

(c) 20 1 1 2 2 ...... n

r r r n r n n rC C C C C C C C C+ + − ++ + + + =

(d)* 2 3 1 11 2 3 ..... ( 1) ( 1)m n m n m n m m nm m m

m m m m mC C C C C C C C n− +− + − + − ⋅ = − ⋅

(e) ( )( ) ( )0 1 1 2 1 1 2( 1)..... .....

!

n

n n nnC C C C C C C C C

n−++ + + =

Q. 9 Find the sums of the following series.

(a) 0 1 2 32 3 4 .....C C C C− + − + (b)* 0 31 2 .....3 4 5 6

C CC C− + − +

Q. 10 Using the binomial theorem, show that

0

rn k n k

r j r jj

C C C−−

=

= ⋅∑Q. 11 Find the sum of the series

0

1 3 7 15( 1) .... terms2 4 8 16

r r rnr n

r r r r rr

C m=

− + + + +

∑

Q. 12* Show that1 2

22 ....... ( 1)n n m nm m m mC C n m C C− +

++ ⋅ + + − + ⋅ =



Section - 4 Section - 1BINOMIAL THEOREM, RATIONAL INDEX

In the previous section, we discussed the expansion of ( )nx y+ , where n is a natural number. We’ll extendthat discussion to a more general scenario now. In particular, we’ll consider the expansion of (1 )nx+ , wheren is a rational number and | x | < 1. Note that any binomial of the form ( )na b+ can be reduced to this form.:

) 1( nn

b aa

a b + =

+ (we are assuming | a | > | b |)

1n

n baa

= +

(1 )n na x= + where | x | < 1

The general binomial theorem states that

2 3( 1) ( 1)( 2)(1 ) 1 ........2! 3!

( 1)( 2)......( 1)........ .......!

n

r

n n n n nx nx x x

n n n n r xr

− − −+ = + + + +

− − − ++ + ∞

That is, there are an infinite number of terms in the expansion with the general term given by

1( 1)( 2).......( 1)

!r

rn n n n rT x

r+− − − +=

For an approximate proof of this expansion, we proceed as follows: assuming that the expansion contains aninfinite number of terms, we have:

( ) 2 30 1 2 31 ... ...n n

nx a a x a x a x a x+ = + + + + + + ∞

Putting x = 0 gives a0 = 1. Now differentiating once gives

( ) 1 21 2 31 2 3 .....nn x a a x a x−+ = + + + ∞

Putting x = 0 gives a1 = n.Proceeding in this way, we find that the rth coefficient is given by

( )( ) ( )1 2 ... 1!n

n n n n ra

r− − − +

=

Note that if n is a natural number, then this expansion reduces to the expansion obtained earlier, because 1rT +

becomes n rrC x , and the expansion terminates for r n> . For the general 1rT + , we obviously cannot use n

rCsince that is defined only for natural n.

One very important point that we are emphasizing again is that the general expansion holds only for | | 1x < .



Let us denote the genral binomial coefficient by Vr. Thus, we have

( 1)( 2)......( 1)!r

n n n n rVr

− − − +=

and0

(1 )n rr

rx V x

∞

=

+ =∑

Let us discuss some particularly interesting expansions. In all cases, | | 1:x <

(1) 1(1 )x −+ : Since 1,n = − we see that

( 1)(( 1) 1)(( 1) 2)......(( 1) 1)!r

rVr

− − − − − − − +=

( 1)r= −

so that the expansion is

1 2 3(1 ) 1 ......x x x x−+ = − + − + ∞

(2) 1(1 )x −− : Again, ( 1)rrV = − and thus

1 2 3(1 ) 1 ......x x x x−− = + + + + ∞

(3) 2(1 )x −+ : We have n = – 2;

( 2)(( 2) 1)(( 2) 2)....(( 2) 1)

!rrV

r− − − − − − − +=

= ( 1) ( 1)!

!

r rr

− +

= ( 1) ( 1)r r− ⋅ +

Thus,

2 2 3(1 ) 1 2 3 4 ......x x x x−+ = − + − + ∞

(4) 2(1 )x −− : Again, ( 1) ( 1)rrV r= − ⋅ + so that

2 2 3(1 ) 1 2 3 4 ........x x x x−− = + + + + ∞

Example – 11

(a) For | | 1x < , expand 3(1 )x −−

(b) Find the coefficient of nx in the expansion of 2 3(1 3 6 10 ....... ) nx x x −+ + + + ∞



Solution: (a) We have

( 3) (( 3) 1)(( 3) 2))......(( 3) 1)

!rrV

r− − − − − − − +=

( 1) ( 2)!

2( !)

r rr

− +=

( 1) ( 1) ( 2)

2

r r r− + +=

Thus,

0 1 2 31, 3, 6, 10........V V V V= = − = = −

so that3 2 3(1 ) 1 3 6 10 .......x x x x−− = + + + + ∞

(b) We use the result of part –(a) in this:

( )2 3 3(1 3 6 10 ....... ) (1 )nnx x x x

−− −+ + + ∞ = −

3(1 ) nx= −

The coefficient of nx in this binomial expansion (note: the power is now a positive integer)

would be 3( 1)n nnC− ⋅ .

Example – 12

Find the magnitude of the greatest term in the expansion of ( ) 2/ 71 5y −− for 18

y = .

Solution: Let us first do the general case: what is the greatest term in the expansion of (1 )nx+ , where n isan arbitrary rational number. We have,

1r

r rT V x+ =

and 11

rr rT V x −

−=

so that1

1

r r

r r

T V xT V

+

−

= ⋅

1n r x

r− += ⋅

Now, let us find the conditions for which this ratio exceeds 1. We have

1r rT T+ ≥

⇒1 11

| |n

r x+ − ≥ ...(1)



For this particular problem, (1) becomes

2 1 17 15

8r

− +− ≥

−

⇒5 817 5r

− ≥

⇒5 13

7 5r≥

⇒2591

r ≤

⇒ r = 0

Thus, 1 1rT T+ = is the greatest term, with magnitude 1.

Example – 13

Find the sum of the series2 2.5 2 5 81 ......6 6.12 6 12 18

⋅ ⋅+ + + + ∞⋅ ⋅

if you are told that this corresponds to an expansion of a binomial, of the form (1 )nx+ .

Solution: We need to determine n and x. For that, we can compare the terms of this series with thecorresponding terms in the following general expansion.

2 3( 1) ( 1) ( 2)(1 ) 1 ......2! 3!

n n n n n nx nx x x− − −+ = + + + + ∞

Thus,

26

nx =

2( 1) 2.52! 6.12

n n x− =

Solving for n and x from these two equations, we get 23

n = − and 12

x = − . Thus, the sum of the

series is

( )2311 1

2nS x

− = + = −

1/34=



Example – 14

Find the sum of the series

1 1 1 ......2 3a b a b a b

+ + ++ + + to n terms

for b << a

Solution: Before solving this problem, ponder a moment over the following fact:

In the expansion of (1 )nx+ , if x << 1, that is, if x is much smaller than 1, then the expansion canbe approximated as

(1 ) 1nx nx+ ≈ +

since all higher order terms can be neglected due to the small magnitude of x.

Coming to the problem, note that if b << a, i.e, if 1ba

<< , then,

1 1

1 ba rb a ra

=+ +

11 1 br

a a

− = +

1 1 rba a

≈ − since 1b

a<<

Thus, the sum S of the series is (to a good approximation)

1

1 1n

r

bS ra a=

≈ − ∑

1 ( 1)

2n n bn

a a+ = −

2

( 1)2

n n n ba a

+= −

Example – 15

Evaluate 993/2 correct to four decimal places.



Solution: We have

( )3/ 23/ 299 100 1= −

( )3/ 23/ 2100 1 0.01= ⋅ −

( )3/ 21000 1 0.01= ⋅ −

( ) ( )2

3 13 2 21000 1 0.01 0.01 ......2 2!

⋅ = ⋅ − ⋅ + ⋅ −

3 31000 1 .......200 80000

= − + −

1000 15 0.0375 ......= − + −

= 985.0375

Note that we only considered the first three terms of the expansion because the higher order termswould not have had any effect on the answer upto the fourth decimal place.



TRY YOURSELF - II

Q. 1 Find the sum of the series

1 1.3 1.3.51 ......3 3.6 3.6.9

+ + + + ∞

Q. 2 Find the sum of the series

1 1.3 1.3.51 ......4 4.8 4.8.12

+ + + + ∞

Q. 3 Find the magnitude of the greatest term in the expansion of ( ) 2/ 51 3y −+ for 15

y = .

Q. 4 Find the magnitude of the greatest term in the expansion of ( )3/ 51 5y− for 13

y =

Q. 5 If | | 1x << , find the approximate value of

3

5

1 (1 )

1 (1 )

x xx x

+ + −

+ + −

Q. 6 If 2 3 4 ......y x x x x= − + − + ∞, show using the general binomial theorem that

1yx

y=

−

Q. 7 If a is very nearly equal to b, then show that the value of 22

b aa b

++ is nearly equal to

13a

b

.

Hint: Write 1a xb

= + , where | | 1x <<

Q. 8 Using the general binomial theorem, find the approximate value for 93/2.

Q. 9* Prove that the coefficient of xn in the expanded represent ation of

2 4

3

1(1 )

x xx

+ −+

will be equal to

( )2 5 81

2n n n + −−

Q. 10 Find the coefficient of xn in the expanded representation of ( ) ( )x

x a x b− − , if | |x < min (a, b)



ASSIGNMENT / BINOMIAL THEOREM

Q. 1 The number of terms in the expansion of ( ) ( )2 1 2 11 1n nx x+ ++ − − is(a) 2n (b) n (c) n + 1 (d) 2n + 1

Q. 2 If the expansion of 3/2

2 n

xx

− has a term independent of x, then n must be of the type

(a) 4 ,k k ∈ I (b) 3 ,k k ∈ I (c) 2 ,k k ∈ I (d) ,k k ∈ I

Q. 3 If the sum of the binomial coefficients in the expansion of 23

2 n

xx

+ is 243, then the term indepen-

dent of x is equal to

(a) 11 (b) 72 (c) 15 (d) 40

Q. 4 The expansion ( ) ( )2 21 1n n

x x x x+ − + − − is a polynomial of degree(a) n (b) 2n − (c) 4n − (d) none of these

Q. 5 Sum of the coefficients in the expansion of ( )103

8022

1 31 3 3x xx x

+ − − + is equal to

(a) –1 (b) 0 (c) 2183 (d) 1

Q. 6 The term which is greatest magnitude in the expansion of 181

2 3x +

when x = –2 is

(a) 11 (b) 12 (c) 13 (d) 7

Q. 7 The greatest coefficient in the expansion of ( )3 21 2 nx −+ is

(a) 3 2 2 11.2

n nnC− −

− (b) 3 2 2 11.2

n nnC− +

+ (c) 3 2 .2n nnC− (d) 3 2 2 2.2n n

nC− −

Q. 8 If A and B respectively denote the sum of the odd terms and sum of the even terms in the expansion of

( )nx y+ , then the value of ( )2 2 nx y− , is equal to

(a) 2 2A B+ (b) 2 2A B− (c) 4AB (d) ( )2A B−

Q. 9 The number of integral terms in the expansion of ( )8732 3+ , is equal to

(a) 23 (b) 7 (c) 15 (d) 11

Q. 10 The coefficient of 83x in ( ) ( ) 32 3 41 1n nx x x x x ++ + + + − , is equal to

(a) ( )7 1 nnC − (b) –

16nC (c) 13

nC (d) 9nC



Q. 11 The value of ( ) ( )0

12 ! 2 2 !

n

r r n r= −∑ , is equal to

(a) ( )

222 !

n

n (b) ( )

2 122 !

n

n

−

(c) 12!

n

n

−

(d) 2

!

n

n

Q. 12 The coefficient of 2x in the expansion of ( )1021 2 3x x+ − is equal to

(a) 65 (b) 210 (c) 90 (d) 150

Q. 13 The value of 1

0 1

nr

r r r

CC C

−

= ++∑ where nr rC C= is equal to

(a) 2nnC (b) 12n− (c)

( )12

n n +(d) 2

n

Q. 14 The value of ( ) ( )( )0

1n

rnr

rC a r b r

=

− − −∑ is equal to

(a) ( )4 nab (b) n na b+ (c) ( )2 2 na b+ (d) 0

Q. 15 The value of expression 0 2 1 3 2 4 2.... n nC C C C C C C C−+ + + + is equal to

(a) 21

nnC − (b) 2

2n

nC − (c) 2nnC (d) none of these

Q. 16 The sum of the last n coefficients in the expansion of ( )2 11 nx −+ when expanding in ascending provesof x is equal to(a) 2 12 n− (b) 2 22 n− (c) 22 n (d) 22 2n n−

Q. 17 The fractional part of ( ) ( )

265

n

n ∈ N is equal to

(a) 0 (b) 1/3 (c) 1/5 (d) 1/6

Q. 18 If the unit digit of 13 7 3 , ,n n n n+ − ∈ N is 7, then the value of n is(a) 4 1,k k+ ∈ I (b) 4 2,k k+ ∈ I (c) 4 3,k k+ ∈ I (d) 4 ,k k ∈ I

Q. 19 The integral value of ( )2 13 1

n++ is equal to

(a) 2 1,k k+ ∈ I (b) ( ) 32 1 ,2

k k+ ∈ I (c) 2 ,k k ∈ I (d) none of these

Q. 20 Let 10 20

0

m

i m ii

S C C −=

= ∑ . The value of m for which S is maximum is

(a) 10 (b) 12 (c) 15 (d) 20



[ TRY YOURSELF - I ]1. 4 43 4+

5. 501301C

9. (a) 0, (b) ( )( )( )( )

22 3 11 2 3

n nn n n

− + ++ + +

11. ( )2 1

2 2 1

mn

mn n

−−

[ TRY YOURSELF - II ]

1. 3

2. 2

3. 1 1T =

4. There are two terms with greatestmagnitude : 1 2 1T T= =

5. 12x+

8. 4.3264

10.1 1 1

n na b b a − −

[ ASSIGNMENT ]

1. c

2. a

3. d

4. a

5. a

6. a

7. a

8. b

9. c

10. b

11. b

12. d

13. d

14. d

15. b

16. b

17. c

18. a

19. c

20. c

ANSWER / BINOMIAL THEOREM

math perm and comb

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