23. Permutation and Combination-2

8/2/2019 23. Permutation and Combination-2

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Mathematics


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Permutation & Combination - 2

Session


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Session Objectives


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Session Objective

1. Combination

2. Circular Permutation


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Combination

Combination | Selection

Selection from a, b, c

Select one

Selection Rejection

a , b , c ,

b , a , c ,c , a , b ,

Select two

Selection Rejection

a , b, c ,

b , c , a ,c , a , b , No. of ways = 3


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Combination

Select threeSelection Rejection

a, b, c J

No. of ways = 1

Number of selection of some from a group.

= Number of rejection of remaining.


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Combination

Number of ways of selecting a group of

two student out of four for a trip to Goa.

S1, S2, S3, S4

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.


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Combination

Number of ways of selecting one group

Of two for Goa other for Agra


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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

! 4 3x.3! P

44 4 43

3 1 (4 3)

P 4!x C C C

3! 1!3! ! ! ! ! !


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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)!p ! !

- n n

r n rC C !


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Questions


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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

11

5

11!(a) C

6!5!!

5

3

6

2

(b) Select 3 man C

Select 2 women C

!

!

5 6

3 2C Cv


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Solution Cont.

(c) At least 3 man

Man 5 Woman - 6

Composition of CommitteeCase Man Woman

3 2 5C3 x6C2 = 150

4 1 5C4 x6C1 = 30

5 0 5C5 x6C0 = 1

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


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In how many ways, a committee of 4

person Can be selected out of 6 personsuch 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 =

5

C3


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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


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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 line

15

2(a) C

(b) 5 point s Collinear

5

2C distinct line Considered as one| p

15 5

2 2Number of Straight line C C 1!


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Find the number of 4 digit numbers

that can be formed by 3 distinct digitsamong 1,2,3,4,5

Solution :- No. of digits = 5

3

1C4!

2!5 3

3 1

4!No. of ways C C

2!@ ! v v

5 digit Select 3 distinct Form 4 digit nos.using these three

3 digit

Select one which

will repeat

Number of digits

formed.

5C3


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In how many ways 9 students can beseated both sides of a table having 5

seat on each side (non-distinguishable)


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Circular Permutation

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

Total line arrangement = 4!

abcd dabc cdab bcda

a

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


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Question


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Solution

(i) Boys 5 Girls 4

Boys (5 1)! 4!! !

Girls 5 4!! v

B1

B2

B3

B4

B5


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Solution Cont.

(ii) Boys 5 Girls 4

B1

B2

B3B4

B5

4!Gs


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Solution Cont.

(iii) Boys 5 Girls 4

B1

B2

B3

B4

B5

2!

G2 G1

G3G4


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Invertible Circular Arrangement

Ex :- Garland, Neck less.

Clockwise and anticlockwise arrangementconsidered as same

ABCD ADCB

(4 1)!

2

|

For n objects.

No. of arrangement =1

(n 1)!2

AC

D

B

C

D

B


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Question


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I have ten different color stones. In

How many way I can make a ring of fivestones

Solution :

Stone - 10

Step 2 :- arrange circularly (Invertible)

Step 1 :- choose 5

10

5C

1 1(5 1)! 4!

2 2 ! v 10

5

1Ans : C 4!

2 v


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Sum of Digits

Find the sum of all three digit numbers

formed by 1, 2 and 3100th 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 repeatedsame no. of times=2=(3-1) !

All digits comes equalno. of times

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


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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 aparticular column) x (No. of 1s as number ofdigits present in the number

Sum of all numbers =


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Class Test


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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?


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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 betogether. 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, no2 boys will be together. So the girls can be seated in 6! (5! + 5!) = 6! 2 5! = 4 5!


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Solution contd..

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

= 4 120 120= 57600

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


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Class Exercise - 2

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


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Solution

Required sum is 3 24 .13.1111 4 .13.111

n n 2

n 1 n 210 1 10 1

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

-

? A24 .13 4444 111!

= 16 13 4333

= 208 4333

= 901264


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Class Exercise - 3

If all the letters of the word SAHARAare arranged as in the dictionary, whatis the 100th word?


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Solution

Arranging the letters alphabetically, we

have A, A, A, H, R, S.5!

602!

!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.


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Class Exercise - 4

How many numbers can be formed

using the digits 3, 4, 5, 6, 5, 4, 3such that the even digits occupythe even places?


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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!

!


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Class Exercise - 5

Ten couples are to be seated around

a table. In how many ways can theybe seated so that no two neighboursare of the same gender?


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Solution

Let all the members of one gender beseated around the table. This can be donein (10 1)! ways. Once one gender isseated, arrangement of other gender is nolonger 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!


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Class Exercise - 6

In how many ways can 15 delegates

be seated around a pentagonal tablehaving 3 chairs at each edge?


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Solution

1

23

4

5

6789

10

11

12

1314

15 15

I2

3

4

5678

9

10

11

12 13

14

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

But we are considering the above two arrangements as samewhile they are clearly different. All that has been done is thatall delegates have shifted one position. One move shift will alsogive a new arrangement.


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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

9

1011

12

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

5


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Class Exercise - 7

Prove that the product of r consecutive

integers is divisible by r!


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Solution

Let 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 rn!

!

rr

n r !n P

n!

! !

But n+rPr is an integer.

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


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Class Exercise - 8

If

find the values of n and r.

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


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Solution

n n

r r 1

p p

!

n! n!

n r ! n r 1 !@ !

n r 1@ !

n n

r 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@ ! !


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Class Exercise - 9

A person wishes to make up as many

parties as he can out of his 18 friendssuch that each party consists of thesame numbers of persons. How manyfriends should he invite?


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Solution

Let the person invite r friends. This canbe done in 18Cr ways. To maximise thenumber of parties, we have to take thelargest value of18Cr. When n is even,nCr will be maximum when r= n/2.

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


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Class Exercise - 10

In how many ways can a cricket team

of 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?


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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 27.6C C 212.1

! ! !


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Class Exercise - 11

In how many ways can we select one

or more items out of a, a, a, b, c, d, e?


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Solution

We can select as in 0 or 1 or 2 or 3, i.e. in 4ways.

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

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

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

@

4

4.2 1 63 ways.! !


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Class Exercise - 12

In how many ways can we divide

10 persons(i) into groups of 5 each,(ii) (ii) into groups of 4, 4 and 2?


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Solution

10

5

C(i) 2!

We have divided by 2!, because if we interchange personsin 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!! v


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Thank you

Documents

23. Permutation and Combination-2