Mathematics

Permutation & Combination - 2

Session

Session Objectives

Session Objective

1. Combination

2. Circular Permutation

Combination

Combination Selection

Selection from a, b, c

Select one

Selection Rejectiona , b , c ,b , a , c ,c , a , b ,

Select two

Selection Rejectiona , b, c ,b , c , a , c , a , b ,

No. of ways = 3

Combination

Select three

Selection Rejectiona, b, c

No. of ways = 1

Number of selection of some from a group.

= Number of rejection of remaining.

Combination

Number of ways of selecting a group of two student out of four for a trip to Goa.

S 1, S 2, S 3, S 4

Select two

Selection Rejection

S1 S2 S3 S4

S1 S3 S2 S4

S1 S4 S2 S3

S2 S3 S1 S4

S2 S4 S1 S3

S3 S4 S1 S2

6 ways.

Combination

Number of ways of selecting one group Of two for Goa other for Agra

Combination

Selection and Arrangement of 3 alphabets from A, B, C, D.

Selection Arrangement Rejection

A, B, C, ABC, ACB, BAC, BCA, CAB, CBA D

A, B, D, ABD, ADB, BAD, BDA, DAB, DBA C

B, C, D, BCD, BDC, CBD, CDB, DBC, DCB A

A, C, D, ACD, ADC, CAD, CDA, DAC, DCA B

43x.3! P

44 4 43

3 1 (4 3)P 4!

x C C C3! 1!3!

Combination

Number of distinct elements = n (1,2,3, .. n).

Ways of rejecting r = nCr

Ways of rejecting rest (n – r) elements = nCn-r

nCr= nCn-r

Particular selection

Total no. of arrangement = nCr.r! = nPr

1,3, … r elements Arrangement r!

nn r

r

P n!C

r! (n r)!r!

n

n r

n! n!r n r C

(n r)!r!n (n r) !(n r)!

n nr n rC C

Questions

Illustrative Problem

There are 5 man and 6 woman. How many way one can select

(a) A committee of 5 person.

(b) A committee of 5 which consist exactly 3 man.

(c) A committee of 5 persons which consist at least 3 man.

Solution :

Man – 5 Woman – 6 Total - 11

115

11!(a) C

6!5!

53

62

(b) Select 3 man C

Select 2 women C

5 6

3 2C C

Solution Cont.

(c) At least – 3 man

Man – 5 Woman - 6

Composition of CommitteeCase Man Woman

3 2 5C3 x 6C2 = 150

4 1 5C4 x 6C1 = 30

5 0 5C5 x 6C0 = 1

No. of Ways = 150 + 30+ 1 = 181.

Illustrative Problem

In how many ways, a committee of 4 person Can be selected out of 6 person such that

(a) Mr. C is always there

(b) If A is there B must be there.

(c) A and B never be together.

Solution :

No. of persons - 6 Committee - 4

(a) Available persons – 5 Persons to select - 3

Available persons – 4Persons to select - 2 Ways = 4C2

(b) Case – 1 : ‘ A is there’ – ‘AB’ in Committee.

Ways = 5C3

Solution Cont.

Case – 2 – ‘A is not there’ – B may /may not be there

Available persons – 5Persons to select - 4 Ways = 5C4

No. of person - 6

Person to Select - 4

(c) ‘AB’ never together = total no. of committee- ‘AB’ always together.

Total no. of committee = 6C4 .

‘AB together in committee’ = 4C2

No. of Ways = 6C4 – 4C2 = 9

Illustrative Problem

How many straight lines can be drawn through 15 given points. when

(a) No. three are collinear

(b) Only five Points are collinear

Solution :

Through two given point and unique straight line15

2(a) C

(b) 5 point s Collinear5

2C distinct line Considered as one 15 5

2 2Number of Straight line C C 1

Illustrative ProblemFind the number of 4 digit numbers that can be formed by 3 distinct digits among 1,2,3,4,5

Solution :- No. of digits = 5

31C 4!

2!5 3

3 1

4!No. of ways C C

2!

5 digit Select 3 distinctForm 4 digit nos. using these three

3 digitSelect one which will repeat

Number of digits formed.

5C3

Illustrative Problem

In how many ways 9 students can be seated both sides of a table having 5 seat on each side (non-distinguishable)

Circular Permutation

A,B,C,D – to be seated in a circular table

Total line arrangement = 4!

abcd dabc cdab bcdaa

b

c

d

1 circular arrangement

4 linear arrangement

4 linear Arrangement 1 circular arrangement

4 ! linear Arrangement

No. of circular arrangement of n object = (n-1) !

4!or (4 1)! circular arrangement.

4

Question

Illustrative Problem

In how many way 4 girls and 5 boys can be seated around a circular table such that

(i) No. two girls sit together

(ii) All girls sit together

(iii) Only two girls sit together

Solution

(i) Boys – 5 Girls – 4

Boys (5 1)! 4!

Girls 5 4!

B1

B2

B3

B4

B5

Solution Cont.

(ii) Boys – 5 Girls – 4

B1

B2

B3B4

B5

4!G’s

Solution Cont.

(iii) Boys – 5 Girls – 4

B1

B2

B3

B4

B5

2!

G2 G1

G3G4

Invertible Circular Arrangement

Ex :- Garland, Neck less.

Clockwise and anticlockwise arrangement considered as same

ABCD ADCB

(4 1)!2

For n objects.

No. of arrangement = 1

(n 1)!2

AC

D

B

C

D

B

Question

Illustrative Problem

I have ten different color stones. In How many way I can make a ring of five stones

Solution :

Stone - 10

Step 2 :- arrange circularly (Invertible)

Step 1 :- choose 510

5C

1 1(5 1)! 4!

2 2 10

5

1Ans : C 4!

2

Sum of Digits

Find the sum of all three digit numbers formed by 1, 2 and 3

100th place 10th place Unit place

1 2 3

1 3 2

2 1 3

2 3 1

3 1 2

3 2 1

12 12 12

Sum of digits same (12)

Each digits is repeated same no. of times=2=(3-1) !

All digits comes equal no. of times

Sum of digits in each column = (1+2+3) x 2! = 12

Sum of Digits

100th place 10th place Unit place

a b c

12 12 12

= 100a +10b +c

= 100x12 +10x12 + 12

= 12 (102 + 10 + 1)

= 12 x 111 = 1332

(sum of all digits) x (No. of repetition in a particular column) x (No. of 1’s as number of digits present in the number

Sum of all numbers =

Class Test

Class Exercise - 1

In how many ways can 5 boys and5 girls be seated in a row so that no2 girls are together and at least2 boys are together?

Solution

First the boys can be seated in 5p5 = 5! ways.

Each arrangement will create six gaps:

__ B __ B __ B __ B __ B

If the girls are seated in the gaps, no 2 girls will be together. Girls can be seated in the gaps in 6p5 = 6! ways. But if the girls are in the first five or the last five gaps, no 2 boys will be together. So the girls can be seated in 6! – (5! + 5!) = 6! – 2 × 5! = 4 × 5!

Solution contd..

Thus, total arrangements possible are 4 x (5!)2

= 4 × 120 × 120= 57600

Note: Under the given condition, more than 2 boys cannot sit together.

Class Exercise - 2

Find the sum of all numbers formedusing the digits 0, 2, 4, 7.

SolutionRequired sum is 3 24 .13.1111 4 .13.111

n n 2n 1 n 2

10 1 10 1n .S. n .S.

10 1 10 1

24 .13 4444 111

= 16 × 13 × 4333

= 208 × 4333

= 901264

Class Exercise - 3

If all the letters of the word ‘SAHARA’are arranged as in the dictionary, whatis the 100th word?

SolutionArranging the letters alphabetically, we have A, A, A, H, R, S.

5!60

2!Number of words starting with A:

Number of words starting with H:5!

203!

Number of words starting with R:5!

203!

Thus, the last word starting with R will be the100th word. This is clearly RSHAAA.

Class Exercise - 4How many numbers can be formedusing the digits 3, 4, 5, 6, 5, 4, 3such that the even digits occupythe even places?

Solution

Even digits are 4, 6, 4.

These can be arranged in the even places in

ways = 3 ways.3!2!

Thus, the total number of ways = 3 × 6 = 18 ways.

The remaining digits: 3, 5, 5, 3 can be arranged

in the remaining places in ways.4!

62! 2!

Class Exercise - 5Ten couples are to be seated arounda table. In how many ways can theybe seated so that no two neighboursare of the same gender?

Solution

Let all the members of one gender be seated around the table. This can be done in (10 – 1)! ways. Once one gender is seated, arrangement of other gender is no longer a problem of circular permutation (since the seats can be identified). Thus, the second gender can be seated in 10! ways.

Thus, total ways = 9! × 10!

Class Exercise - 6In how many ways can 15 delegatesbe seated around a pentagonal tablehaving 3 chairs at each edge?

Solution

12

3

4

5

6789

10

11

12 13

14 15 15

I2

3

4

5678

9

10

1112

1314

If we consider the problem as one of circular permutations, the answer is (15 – 1)! = 14!

But we are considering the above two arrangements as same while they are clearly different. All that has been done is that all delegates have shifted one position. One move shift will also give a new arrangement.

Solution contd..

Thus, we are counting three different arrangements as one.

Thus, number of actual arrangements possible = 3 × 14!

But after three shifts, the arrangement will be

which is identical to the original arrangement.

1314

15

1

2

3456

7

8

910

1112

or , where 5 is the number of sides of regular polygon.15!5

Class Exercise - 7Prove that the product of r consecutiveintegers is divisible by r!

SolutionLet the r consecutive integers be

(n + 1), (n + 2), (n + 3), ..., (n + r)

Product = (n + 1)(n + 2)(n + 3) ... (n + r)

n! n 1 n 2 n 3 ... n r

n!

rr

n r !n P

n!

But n+rPr is an integer.

Thus, the product of r consecutive integers is divisible by r!(Proved)

Class Exercise - 8If

find the values of n and r.

n n n nr r 1 r r 1p p and C C ,

Solutionn n

r r 1p p

n! n!

n r ! n r 1 !

n r 1

n nr r 1Also C C

n! n!

n r ! r! n r 1 ! r 1 !

n! n!

n r ! r! n r 1 ! r 1 !

r 2, n 3

Class Exercise - 9A person wishes to make up as manyparties as he can out of his 18 friendssuch that each party consists of thesame numbers of persons. How manyfriends should he invite?

Solution

Let the person invite r friends. This canbe done in 18Cr ways. To maximise thenumber of parties, we have to take thelargest value of 18Cr. When n is even,nCr will be maximum when r= n/2.

Thus, he should invite 18/2 = 9 friends.

Class Exercise - 10In how many ways can a cricket teamof 5 batsmen, 3 all-rounders, 2 bowlersand 1 wicket keeper be selected from19 players including 7 batsmen,6 all-rounders, 3 bowlers and3 wicket keepers?

Solution

The batsmen can be selected in 7C5 = 21 ways.

The all-rounders can be selected in 6C3 = 20 ways.

The bowlers can be selected in 3C2 = 3 ways.

The wicketkeeper can be selected in 3C1 = 3 ways.

Thus, the total ways of selecting the team

= 21 × 20 × 3 × 3 = 3780 ways.

Note: 7 75 2

7.6C C 21

2.1

Class Exercise - 11In how many ways can we select oneor more items out of a, a, a, b, c, d, e?

Solution

We can select ‘a’s in 0 or 1 or 2 or 3, i.e. in 4 ways.

We can select ‘b’ in 2 ways, i.e. either we select it or we do not select it and so on.

The required number of ways in which we can select one or more items is

(3 + 1)(1 + 1)(1 + 1)(1 + 1)(1 + 1) – 1

44.2 1 63 ways.

Class Exercise - 12

In how many ways can we divide10 persons (i) into groups of 5 each,(ii) (ii) into groups of 4, 4 and 2?

Solution

105C

(i)2!

We have divided by 2!, because if we interchange persons in group one, with persons in group two, the division is not different, i.e.

group 1 group 2 group 2 group 1

abdfj ceghi ceghi abdfj

10 6 24 4 2C . C . C 10! 1

(ii)2! 4! 4! 2! 2!

Thank you