permutation 2

7/25/2019 Permutation 2

1/94

Permutations and

Combinations

Quantitative Aptitude Business Statistics


2/94

Quantitative Aptitude & Business

Statistics:Permutations andCombinations2

The Fundamental Principle of

Multiplication

If there are

n1 ways of doing one operation,

n2 ways of doing a second operation,n3 ways of doing a third operation ,

and so forth,


3/94



then the sequence of k operations

can be performed in n1 n2 n3.. nkways.

N= n1 n2 n3.. nk


4/94



Example 1 A used car wholesaler has agents who

classify cars by size (full, medium, and

compact) and age (0 - 2 years, 2- 4

years, 4 - 6 years, and over 6 years).

Determine the number of possible

automobile classifications.


5/94



Solution

Full(F)

Compact

(C)

Medium

(M)

0-2

2-4

4-6

>6

0-2

2-4

4-6

>6

0-2

2-4

4-6

>6

The tree diagram enumerates all possible

classifications, the total number of which

is 3x4= 12.


6/94



Example 2

Mr. X has 2 pairs of trousers, 3 shirts

and 2 ties.

He chooses a pair of trousers, a shirt

and a tie to wear everyday. Find the maximum number of days

he does not need to repeat his

clothing.


7/94



Solution

The maximum number of days he

does not need to repeat his clothing

is 232 = 12


8/94



1.2 Factorials

The product of the first nconsecutive integers is denoted by n!

and is read as factorial n.

That is n! = 1

2

3

4

. (n-1) n For example,

4!=1x2x3x4=24,

7!=1234567=5040.

Note 0! defined to be 1.


9/94



The product of any number of

consecutive integers can be

expressed as a quotient of twofactorials, for example,

6789 = 9!/5! = 9! / (9 4)!

1112131415= 15! / 10!

=15! / (15 5)!

In particular,

n(n 1)(n 2)...(n r + 1)

=


10/94



1.3 Permutations

(A) Permutations A permutation is an arrangement of

objects.

abc and bca are two differentpermutations.


11/94



1. Permutations with repetition

The number of permutations of r objects,

taken from n unlike objects,

can be found by considering the number ofways of filling r blank spaces in order with

the n given objects.

If repetition is allowed, each blank space

can be filled by the objects in n differentways.


12/94



Therefore, the number of

permutations of r objects, takenfrom n unlike objects,

each of which may be repeatedany number of times

= n n n .... n(r factors)= nr

n n n n n

1 2 3 4 r


13/94



2. Permutations without repetition

If repetition is not allowed, thenumber of ways of filling eachblank space is one less than thepreceding one.

n n-1 n-2 n-3 n-r+1

1 2 3 4 r


14/94



Therefore, the number of permutations ofr objects, taken from n unlike objects,

each of which can only be used once in

each permutation

=n(n 1)(n2) .... (nr + 1)

Various notations are used to represent

the number of permutations of a set ofn elements taken r at a time;


15/94



some of them are

),(,, rnPPPrn

n

r ),(,, rnPPP

rn

n

r ),(,, rnPPP

rn

n

r ),(,, rnPPP

rn

n

r ),(,, rnPPP

rn

n

r ),(,, rnPPP

rn

n

r

),(,, rnPPPrn

n

r

n

rP

rnnnn

rn

rnrnnnn

rn

n

=

+=

+=

)1)....(2)(1(123)...(

123)...)(1)....(2)(1(

)!(!

Since

We have

)!(

!

rn

nP

n

r

=


16/94



Example 3 How many 4-digit numbers can be made

from the figures 1, 2, 3, 4, 5, 6, 7 when

(a) repetitions are allowed;

(b) repetition is not allowed?


17/94



Solution

(a) Number of 4-digit numbers= 74= 2401.

(b) Number of 4 digit numbers

=7 6 5 4= 840.


18/94



Example 4

In how many ways can 10 men bearranged

(a) in a row,

(b) in a circle?

Solution

(a) Number of ways is

= 3628800

10

10P

S


19/94



Suppose we arrange

the 4 letters A, B, C

and D in a circular

arrangement as shown. Note that the

arrangements ABCD,

BCDA, CDAB and DABC

are not distinguishable.

A

D

C

B


20/94



For each circular arrangement thereare 4 distinguishable arrangements

on a line. If there are P circular arrangements,

these yield 4P arrangements on aline, which we know is 4!.

!3)!14(4

!4===PHence


21/94


Statistics:Permutations and

Combinations21

The number of distinct circular

arrangements of n objects is(n1)!

Hence 10 men can be arranged

in a circle in 9! = 362 880 ways.

Solution (b)


22/94



Combinations22

(B) Conditional Permutations When arranging elements in order ,

certain restrictions may apply.

In such cases the restriction shouldbe dealt with first..


23/94



Combinations23

Example 5How many even numerals between200 and 400 can be formed by using1, 2, 3, 4, 5 as digits

(a) if any digit may be repeated;(b) if no digit may be repeated?


24/94



Combinations24

Solution (a)

Number of ways of choosing the

hundreds digit = 2. Number of ways of choosing the tens

digit = 5.

Number of ways of choosing the unit

digit = 2.

Number of even numerals between

200 and 400 is

2 5 2 = 20.


25/94



Combinations25

Solution (b)

If the hundreds digit is 2,

then the number of ways of choosingan even unit digit = 1,

and the number of ways of choosing a

tens digit = 3.

the number of numerals formed

113 = 3.


26/94



Combinations26

If the hundreds digit is 3, then the

number of ways of choosing aneven. unit digit = 2, and the

number of ways of choosing a tens

digit = 3.

number of numerals formed

= 1

2

3 = 6.

the number of even numeralsbetween 200 and 400 = 3 + 6 =

9


27/94



Combinations27

Example 6In how many ways can 7different books be

arranged on a shelf(a) if two particular booksare together;


28/94



Combinations28

Solution (a)

If two particular books are together,

they can be considered as one book for

arranging.

The number of arrangement of 6 books

= 6! = 720.

The two particular books can bearranged in 2 ways among themselves.

The number of arrangement of 7 books

with two particular books together= 6! x 2 = 1440.


29/94



Combinations29

(b) if two particular books areseparated?

Solution (b)

Total number of arrangement of 7books = 7! = 5040.

the number of arrangement of 7

books with 2 particular books

separated = 5040 1440 = 3600.


30/94



Combinations30

(C) Permutation with

Indistinguishable Elements

In some sets of elements there may

be certain members that are

indistinguishable from each other.

The example below illustrates how

to find the number of permutations in

this kind of situation.


31/94


32/94



Combinations32

However, the 3 Ss are

indistinguishable from each other and

can be permuted in 3! different ways.

As a result, each of the 9!

arrangements of the letters of

ISOSCELES that would otherwisespell a new word will be repeated 3!

times.


33/94



Combinations33

To avoid counting repetitions resulting

from the 3 Ss, we must divide 9! by 3!.

Similarly, we must divide by 2! to avoid

counting repetitions resulting from the

2 indistinguishable Es.

Hence the total number of words thatcan be formed is

9! 3! 2! = 30240


34/94



Combinations34

If a set of n elements has k1

indistinguishable elements of one kind,

k2of another kind,

and so on for r kinds of elements, then

the number of permutations of the set of

n elements is

!!!

!

21 rkkk

n


35/94



Combinations35

1.4 Combinations

When a selection of objects is madewith no regard being paid to order, it is

referred to as a combination.

Thus, ABC, ACB, BAG, BCA, CAB, CBA

are different permutation, but they arethe same combination of letters.


36/94



Combinations36

Suppose we wish to appoint a

committee of 3 from a class of 30students.

We know that P330 is the number of

different ordered sets of 3 studentseach that may be selected from among

30 students.

However, the ordering of the students

on the committee has no significance,


37/94



Combinations37

so our problem is to determine the

number of three-element unordered

subsets that can be constructed

from a set of 30 elements.

Any three-element set may be

ordered in 3! different ways, so P330is 3! times too large.

Hence, if we divide P330 by 3!,the

result will be the number of

unordered subsets of 30 elements

taken 3 at a time.


38/94



Combinations38

This number of unordered subsets is

also called the number of

combinations of 30 elements taken 3at a time, denoted by C330 and

4060!3!27

!30!3

1 303

30

3

==

= PC


39/94



Combinations39

In general, each unordered r-

element subset of a given n-element

set (r n) is called a combination. The number of combinations of n

elements taken r at a time is

denoted by Cnror nCr or C(n, r) .


40/94



Combinations40

A general equation relating

combinations to permutationsis

!)!(

!

!

1

rrn

nP

rC

n

r

n

r

==


41/94


42/94



Combinations42

Example8

If 167 C 90+167 C x =168 C xthen x is

Solution: nCr-1+nCr=n+1 Cr Given 167 C90+167cx=168C x

We may write

167C91-1

+ 167 C91

=167+1 C61

=168 C91

X=91


43/94



Combinations43

Example9

If 20 C 3r= 20C 2r+5,find r

Using nCr=nC n-r in the right side ofthe given equation ,we find ,

20 C 3r =20 C 20-(2r+5)

3r=15-2r

r=3


44/94



Combinations44

Example 10

If 100 C 98=999 C 97 +x C 901 find x.

Solution 100C 98=999C 98+999C97 = 999C901+999C97

X=999


45/94


46/94



Combinations46

Example12

If n C r-1=36 ,n Cr =84 and n C r+1 =126 thenfind r

Solution

n-r+1 =7/3 * r

3/2 (r+1)+1 =7/3 * r

r=3

3

7

36

84

1

==

r

r

nC

nC

2

3

84

1261==

+

r

r

nC

nC


47/94



Combinations47

Example 13 How many different 5-card hands

can be dealt from a deck of 52

playing cards?


48/94



Combinations48

Solution

Since we are not concerned with the

order in which each card is dealt,

our problem concerns the number of

combinations of 52 elements taken

5 at a time.

The number of different hands isC525 2118760.


49/94



Combinations49

Example 14

6 points are given and no three of

them are collinear.(a) How many triangles can be

formed by using 3 of the given points

as vertices?


50/94



Combinations50

Solution:

Solution

(a) Number of triangles = number of ways

of selecting 3 points out of 6

= C63

= 20.


51/94



Combinations51

b) How many pairs of triangles can beformed by using the 6 points as

vertices ?


52/94



Combinations52

Let the points be A, B, C, D, E, F. If A, B, C are selected to form a

triangles, then D, E, F must form theother triangle.

Similarly, if D, E, F are selected toform a triangle, then A, B, C must

form the other triangle.


53/94


54/94



Combinations54

Example 15 From among 25 boys who play

basketball, in how many different ways

can a team of 5 players be selected ifone of the players is to be designated

as captain?


55/94



Combinations55

Solution

A captain may be chosen from any of the

25 players.

The remaining 4 players can be chosen inC254

different ways.

By the fundamental counting principle, the

total number of different teams that can

be formed is25 C244265650.


56/94



Combinations56

(B) Conditional Combinations

If a selection is to be restricted in

some way, this restriction must bedealt with first.

The following examples illustrate

such conditional combination

problems.


57/94



Combinations57

Example 16

A committee of 3 men and 4women is to be selectedfrom 6 men and 9 women.

If there is a married coupleamong the 15 persons, inhow many ways can thecommittee be selected sothat it contains the marriedcouple?

Solution


58/94



Combinations58

Solution

If the committee contains the married

couple, then only 2 men and 3 women

are to be selected from the remaining 5

men and 8 women.

The number of ways of selecting 2 men

out of 5 = C52= 10.


59/94



Combinations59

The number of ways of selecting 3

women out of 8 =C83= 56.

the number of ways of selecting the

committee = lO 56 = 560.


60/94



Combinations60

Example 17 Find the number of ways a team of 4

can be chosen from 15 boys and 10

girls if(a) it must contain 2 boys and 2 girls,


61/94



Combinations61

Solution (a)

Boys can be chosen in C152= 105

ways

Girls can be chosen in C102= 45

ways.

Total number of ways is 105 45 =4725.

(b) it must contain at least 1 boy and


62/94



Combinations62

(b) t ust co ta at east boy a d

1 girl.

Solution : If the team must contain at least 1 boy

and 1 girl it can be formed in thefollowing ways:

(I) 1 boy and 3 girls, with C151 C10

3=

1800 ways,

(ii) 2 boys and 2 girls, with 4725 ways,

(iii) 3 boys and 1 girl, with C15

3 C10

1

=4550 ways.

the total number of teams is

1800 + 4725 + 4550 = 11075.


63/94



Combinations63

Example 18 Mr. .X has 12 friends and wishes to

invite 6 of them to a party. Find thenumber

of ways he may do this if

(a) there is no restriction on choice,


64/94



Combinations64

Solution (a)

An unrestricted choice of 6out of 12 gives C126 924.


65/94



Combinations65

(b) two of the friends is a couple andwill not attend separately,


66/94



Combinations66

B Solution

If the couple attend, the remaining 4

may then be chosen from the other

10 in C104ways.

If the couple does not attend, then

He simply chooses 6 from the other

10 in C10

6

ways. total number of ways is C104 + C

106

= 420.


67/94



Combinations67

Example 19

Find the number of ways inwhich 30 students can bedivided into three groups, eachof 10 students, if the order ofthe groups and thearrangement of the students in

a group are immaterial.

Solution


68/94



Combinations68

Let the groups be denoted by A, Band C. Since the arrangement of thestudents in a group is immaterial,

group A can be selected from the 30students in C3010ways .


69/94



Combinations69

Group B can be selected from the

remaining 20 students in C2010ways.

There is only 1 way of forming group

C from the remaining 10 students.


70/94



Combinations70

Since the order of the groups is

immaterial, we have to divide theproduct C3010 C

2010 C

1010by 3!,

hence the total number of ways of

forming the three groups is

10

10

20

10

30

3!31 CCC


71/94



Combinations71

Example20

If n Pr = 604800 10 C r =120 ,findthe value of r

We Know that nC r .r P r = nPr .

We will use this equality to find r

10Pr =10Cr .r| r |=604800/120=5040=7 |

r=7


72/94



Combinations72

Example 21

Find the value of n and r

n Pr= n P r+1and

n Cr= n C r-1

Solution : Givenn Pr= n P r+1

n r=1 (i)

n Cr= n C

r-1n-r = r-1 (ii)

Solving i and ii

r=2 and n=3


73/94



Combinations73

Multiple choice Questions


74/94



Combinations74

1. Eleven students are participating

in a race. In how many ways the

first 5 prizes can be won?

A) 44550

B) 55440

C) 120

D) 90


75/94



Combinations75

1. Eleven students are participating

in a race. In how many ways the

first 5 prizes can be won?

A) 44550

B) 55440

C) 120

D) 90


76/94



Combinations76

2. There are 10 trains plying between

Calcutta and Delhi. The number of ways in

which a person can go from Calcutta toDelhi and return

A)99.

B) 90

C) 80 D) None of these


77/94



Combinations77

2. There are 10 trains plying between

Calcutta and Delhi. The number of ways in

which a person can go from Calcutta toDelhi and return

A)99.

B) 90

C) 80 D) None of these


78/94



Combinations78

3. 4P4 is equal to

A) 1

B) 24

C) 0

D) None of these


79/94



Combinations79

3. 4P4 is equal to

A) 1

B) 24

C) 0

D) None of these


80/94



Combinations80

4.In how many ways can 8 persons

be seated at a round table?

A) 5040

B) 4050

C) 450

D) 540


81/94



Combinations81

4.In how many ways can 8 persons

be seated at a round table?

A) 5040

B) 4050

C) 450

D) 540


82/94



Combinations82

5. If then

value of n is

A) 15

B) 14

C) 13

D) 12

n n+113 12P : P =3: 4


83/94



Combinations83

5. If then

value of n is

A) 15

B) 14

C) 13

D) 12

n n+113 12

P : P =3: 4


84/94



Combinations84

6.Find r if 5Pr = 60

A) 4

B) 3

C) 6

D) 7


85/94



Combinations85

6.Find r if 5Pr = 60

A) 4

B) 3

C) 6

D) 7


86/94



Combinations86

7. In how many different ways can

seven persons stand in a line for a

group photograph?

A) 5040

B) 720

C) 120 D) 27


87/94



Combinations87

7. In how many different ways can

seven persons stand in a line for a

group photograph?

A) 5040

B) 720

C) 120 D) 27


88/94



Combinations88

8. If then the value of

n is ______

A)0

B) 2

C) 8

D) None of above

18 18

n n+2C = C


89/94



Combinations89

8. If then the value of

n is ______

A)0

B) 2

C) 8

D) None of above

18 18

n n+2C = C


90/94



Combinations90

9. The ways of selecting 4 letters

from the word EXAMINATION is

A) 136.

B) 130

C) 125

D) None of these


91/94



Combinations91

9. The ways of selecting 4 letters

from the word EXAMINATION is

A) 136.

B) 130

C) 125

D) None of these


92/94



Combinations92

10 If 5Pr = 120, then the value of r

is

A) 4,5

B) 2

C) 4

D) None of these


93/94



Combinations93

10 If 5Pr = 120, then the value of r

is

A) 4,5

B) 2

C) 4

D) None of these


94/94

THE END

Permutations and

Combinations

permutation 2

Documents