22. Permutation and Combination-1

7/29/2019 22. Permutation and Combination-1

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Mathematics


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Permutation & Combination

Session


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Session Objective

1. Factorial2. Fundamental principles of counting

3. Permutations as arrangement

4.n

Pr formula5. Permutations under conditions

6. Permutation of n objects taken r at a time


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Permutation Its arrangement

Two element a b

Arrangements (a, b), (b, a) = 2

FatherSon

Father is riding

FatherSon

Father is teaching


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Three elements a, b, c

Arrangements :

a b c

a c b

b a c

b c a

c a b

c b a

= 6

First Second ThirdPlace Place Place


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Ist 2nd 3rd

3 ways 2 ways 1 ways

Total ways = 3 + 2 + 1 ?

or

3 x 2 x 1 ?


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Ist 2nd

2 ways 1 ways

xTotal ways = 2 + 1 = 3or

2 x 1 = 2


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Ist 2nd 3rd

3 ways 2 ways 1 ways

xTotal ways = 3 + 2 + 1 ?or

3 x 2 x 1 ?


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Number of Modes A B?

Cycle

Scooter

Car

BA1 Km

Number of modes to reach B =

Add/Multiply

1 1 1 1

Bus

Scooter

Car Walking

Number of ways A B?

I

IIBA No. of ways = 1 + 1 = 2

Independent Process

+ + + = 4


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Mode & Way

Number of style A B?

I

IIBA

Scooter

Cycle

CarWays 2Modes 4

To reach B, one dependent on both ways and mode.

Number of style = 4 x 2 = 8

Independent Process +Dependent Process X


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Mode & Way

There are two ways and 4 modes for

A B. How many way one can reachB from A?

One can reach Lucknow from New Delhionly through Kanpur (No direct root)

I

II

III

IV

A

B

Kanpur LucknowNewDelhi

Process Dependent

I AI BII AII BIII A

III BIV AIV B

No. of ways = 4 x 2 = 8


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Mode & Way

IIIIIIIV

A

BKanpur

LucknowNewDelhi

IV

V


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Questions


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Illustrative Problem

From the digits 1, 2, 3, 4, 5 how many two

digit even and odd numbers can be formed.Repetition of digits is allowed.

Solution: Total nos = 5

Even number

5 ways 2 ways (2/4)

Even numbers=5 x 2=10


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Solution contd..

Odd number

5 ways 3 ways (1/3/5)

Odd numbers = 5 x 3 = 15

Total numbers

5 ways 5 ways

Total numbers = 5 x 5 = 25

Even Numbers=10


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From the digits 1, 2, 3, 4, 5 how many two

digit numbers can be formed. Whenrepetition is not allowed.

Solution:

5 ways 4 ways

Total = 5 x 4 = 20


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There are three questions. Every question can

be answered in two ways, (True or False). Inhow many way one can answer these threequestions?

Solution:

2 ways(T/F)

2 ways

Question Ist 2nd 3rd

2 ways

No. of ways = 2 x 2 x 2 = 8

T

T

F

T

FT

F

F

T

F

T

FT

F


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In a class there are 10 boys and 8 girls.

For a quiz competition, a teacher howmany way can select(i) One student(ii) One boy and one girl student

(i) Independent of whether boy/girl = 10 + 8 = 18ways

Solution:

(ii)Dependent process = 10 x 8 = 80 ways

8ways 10 ways

Girl Boy


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8 10

Solution : (iii) Girl Boy1 Boy2

(iv)student=10+8=18

In a class there are 10 boys and 8 girls.For a quiz competition, a teacher how

many way can select(iii) two boys and one girl(iv) two students

9 10 x 9 x 8 = 720


=18 x 17

18 17

student1 student2


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In a class there are 10 boys and 8 girls.For a quiz competition, a teacher how

many way can select(v) at least one girl while selecting 3students

Solution: case1: 1 girl

8 10 9 10 x 9 x 8 = 720

boys-10girl-8

G B B

case2: 2 girl

8 7 108 x 7 x 10 = 560

G G B

case3: 3 girl 8 x7 x 6 = 186

Ans: 720+560+186=1666



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Principle of Counting

Multiplication Principle : If a job can be

done in m different ways, following whichanother can be done in n different ways andso on. Then total of ways doing the jobs= m x n x ways.

Addition Principle : If a job can bedone in m different ways or n different waysthen number of ways of doing the job is (m + n).

Multiplication

Dependent ProcessAddition Independent Process


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Questions


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Eight children are to be seated on a bench.How many arrangements are possible if the

youngest and eldest child sits at left and rightcorner respectively.

Solution:

We have 6 children to be seated

6 5

Youngest Eldest

4 3 2 1

No. ways = 6 x 5 x 4 x 3 x 2 x 1 = 720 x 2 = 1440 ways


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A class consists of 6 girls and 8 boys. In how

many ways can a president, vice president,treasurer and secretary be chosen so thatthe treasurer must be a girl and secretarymust be a boy. (Given that a student canthold more than one position)

Solution :

Girls 6Boys - 8

Treasurer (Girl)6 ways

Girls 5Boys - 8

Secretary (Boy)8 ways


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Solution contd..

Girls 5Boys 7;Total = 12

12 ways

President Vice President

11 ways

Total = 6 x 8 x 12 x 11

Treasurer(Girl)-6 ways

Secretary(Boy)-8 ways


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Factorial

Defined only for non-negative integers

Denoted as n ! or .n

N ! = n . (n - 1) . (n 2) 3 . 2 . 1.

Special case : 0 ! = 1

Example :

3 ! = 3 . 2 . 1 = 65 ! = 5 . 4 . 4 . 3. 2 . 1 = 120(4.5) ! - Not defined(-2) ! - Not defined


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Questions


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1 1 x

If then find x.9! 10! 11! =

Solution :

110 11 121= =

11! 11!x9! 10!

=

11 10 9! 11 10!

9! 10!

=


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Find x if 8! x! = (x + 2)! 6!

Solution :

(x 2)! 8!

x! 6!

=

(x 2)(x 1)x! 8.7.6!

x! 6!

=

(x 2)(x 1) 8.7 56 = =

Shortcut

2

2

x 3x 2 56

x 3x 54 0

=

= x 9, 6

As x 0 x 6

=

=


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Permutation

Arrangement of a number of object(s) taken

some or all at a fine.

Example :Arrangement of 3 elements out of

5 distinct elements = 5.4.3.2.1 5!5.4.3 2.1 2!= =

5

3

5!P

(5 3)!= =


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nPr

Arrangement of r elements out of n givendistinct elements.

n (n - 1)

1st 2nd rth

(n r + 1)

?

.

No. of arrangement

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

=


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nPr

n.(n 1)(n 2)....(n r 1)(n r)....2.1

(n r)(n r 1)...2.1

=

No. of arrangement

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

=

?!

?!=

n

r

n!P (r n)

(n r)!= =

For r = n

Arrangements of n distinct element nPn = n!taken all at a time.


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Questions


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In how many way 4 people (A, B, C, D)

can be seated

(a) in a row(b) such that Mr. A and Mr. B always sit

together

Solution :

(a) 4P4 = 4! (b) (A, B),

treat as one

C, D


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Solution contd..

Arrangement among 3 = 3P3

(A, B, C, D)

(B, A, C, D)2P2

(A, B, D, C)

(B, A, D, C)2P2

3P3

(A, B), C, D

(A, B), D, C

C, (A, B), D

C, D, (A, B)D, (A, B), C

D, C, (A, B)

3P3 x2P2

(b) (A, B), C, D

= 3!2! = 12


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Solution contd..

In how many way 4 people (A, B, C, D)

can be seated(c) A,B never sit together

= Total no. of arrangement No. (A, B) together

= 4P43P3.

2P2= 4! - 3! 2!= 24 12 = 12


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Seven songs (Duration 4, 4, 5, 6, 7,

7, 7, 7 mins.) are to be rendered in aprogramme(a) How many way it can be done(b) such that it occurs in ascending order

(duration wise)

Solution :

(a) 7P7 = 7!

(b) Order 4, 4, 5, 6, 7, 7, 7 mins2P2

3P3

No. of way = 2P2 x3P3 = 2! 3! = 24


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How many four digits number can be

formed by the digits. 3, 4, 5, 6, 7, 8such that

(a) 3 must come(b) 3 never comes

(c) 3 will be first digit(d) 3 must be there but not first digit


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Solution

(a) 3, 4, 5, 6, 7, 8

3

33

3

5

55

5

3

33

3

5

3

5

3

5

3

5

3

P

PP

P

Digitsavailable

Position Arrangements

No. of 4 digit numbers with 3 = 4 x 5P3

3 can take any of four position.In each cases. 5 digits to be arranged in 3 position.


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Solution contd..

(b) Digits available 5 (4, 5, 6, 7, 8

No. of 4 digit numbers without 3 = 5P4

(c) 3 _ _ _

No. of digits available = 5No. of position available = 3

No. of 4 digit number start with 3 =5

P3


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Solution contd..

(d) 4 digit nos. contain 3 but not at first= 4 digit number with 3 4 digit

number with 3 at first

= solution (a) solution (c)

= 4.5

P35

P3 = 3.5

P3


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In how many way a group photograph of

7 people out of 10 people can be taken.Such that(a) three particular person always be there(b) three particular person never be there(c) three particular always be together

Solution:

(a)3 particular 7 places 7P3

With each arrangement

X _ X _ X _ _

Arrangements


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Solution contd..

Arrangements

Person available 7Places available 3

7P4

Total no. of arrangements = 7P3 x7P4

Person available = 7Places available = 7

7P7arrangements


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Solution contd..

(c) X X X _ _ _ _

Treat as one

No. of person = 10 7 + 1 = 8

Place available = 5of which one (3 in 1) always be there.

No. of arrangement = 5.7P4

3 Particular can be arranged =3

P3 way

Total arrangement = 5.7P4.3P3


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How many way, 3 chemistry, 2 physics,

4 mathematics book can be arranged suchthat all books of same subjects are kepttogether.

Solution:

1 2 3 1 2 1 2 3 4

Chemistry Physics Mathematics

Intra subject

Arrangements 3P32P2

4P4


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Solution contd..

Inter subject arrangement

Phy Chem Maths Arrangement = 3P3

Total no. of arrangements = 3P3(3p3 x2p2 x

4p4)


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Class Test


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Class Exercise - 1

If are in the

ratio 2 : 1, find the value of n.

n! n!and

2! n 2 ! 4! n 4 !


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Solution

Given that

4! n 4 !n! 2

2! n 2 ! n!

=

4 3 2! n 4 !2

2! n 2 n 3 n 4 !

=

4 32

n 2 n 3

=

2n 5n 6 6 = 2n 5n 0 = n n 5 0 =

n 0 or n 5. = =

Answer is n = 5 (rejecting n = 0).


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Class Exercise - 2

How many numbers are there between100 and 1000 such that each digit iseither 3 or 7?


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Solution

By fundamental principle of counting,

the required number = 2 2 2 = 8(Each place has two choices.)


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Class Exercise - 3

How many three-digit numbers can beused using 0, 1, 2, 3 and 4, if

(i) repetition is not allowed, and(ii) repetition is allowed?


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Solution

(i) Hundreds place can be filled in 4 ways.

Tens place can be filled in 4 ways.

Units place can be filled in 3 ways.

Required number = 4 4 3 = 48

(ii) Similarly, the required number = 4 5 5 = 100


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Class Exercise - 4

How many four-digit numbers have atleast one digit repeated?


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Solution

Number of four-digit numbers

= 9 10 10 10 = 9000

Number of four-digit numbers with no repetition= 9 9 8 7 = 4536

Number of four-digit numbers with at leastone digit repeated = 9000 4536 = 4464


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Class Exercise - 5

There are 5 periods in a school and 6subjects. In how many ways can the

time table be drawn for a day so thatno subject is repeated?


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Solution

Six subjects can be allocated to five periods inways, without a subject being repeated.

6

9p 6!=


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Class Exercise - 6

Number of ways in which 7 differentsweets can be distributed amongst5 children so that each may receiveat most 7 sweets is

(a) 75 (b) 57

(c)7

p5 (d) 35


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Solution

Each sweet can be given to any of the 5

children.

Thus, the required number is

5 5 5 5 5 5 5 = 57

Hence answer is (b)


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Class Exercise - 7

In how many ways can 5 studentsbe seated such that Ram alwaysoccupies a corner seat and Seetaand Geeta are always together?


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Solution

Seeta and Geeta can be arranged in 2 ways.

Remaining students can be arranged in 2ways.

Total ways = 2 3 2 2 = 24

Ram can be seated in 2 ways. Seetaand Geeta can be together in 3 ways.(If Ram occupies seat 1, Seeta-Geetacan be in 2-3, 3-4 or 4-5.)


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Class Exercise - 8

How many words can be made fromthe word helicopter so that thevowels come together?


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Solution

Treating the vowels as one unit, we have 7 units.

These can be arranged in 7! ways.

The vowels can be arranged in 4! ways.

Total ways = 7! 4! = 120960

l


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Class Exercise - 9

There are 5 questions. Each question has

two options (one answer is correct). Inhow many ways can a student fill up theanswer sheet, when he is asked toattempt all the questions?

S l i


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Solution

Every question can be answered in two ways.

= 2 2 2 2 2 = 25 = 32 ways.

Five questions can be answered in

Cl E i 10


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Class Exercise - 10

In how many ways can 6 students (3 boysand 3 girls) be seated so that

(i) the boys and girls sit alternatively,(ii) no 2 girls are adjacent?

S l ti


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Solution

(i) Case 1: From left-side, when first student isa boy, then the boys can occupy Ist, 3rd and

5th places. And the girls can occupy 2nd, 4thand 6th places.

So the boys can be seated in 3p3 ways and the

girls can be seated in 5p3 ways.

Number of arrangement = 3! . 3!

S l ti td


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Solution contd..

Case 2: When the first student is a girl(from left), then also the number of permutation = 3! 3!

Therefore, total number of permutation = 2 (3!)2 = 72

(ii) Let first boys are seated.

They can sit in three places in 3p3 = 3! ways. Since no girlsshould be adjacent, the number of seats left for girls are four.

__ B __ B __ B __

Number of permutation for girls = 4p3 = 4!

Therefore, total number of permutation = 3! 4! = 144


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Thank you

22. Permutation and Combination-1

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