mathematics. permutation & combination session

Mathematics

Permutation & Combination

Session

Session Objective

1. Factorial

2. Fundamental principles of counting

3. Permutations as arrangement

4. nPr formula

5. Permutations under conditions

6. Permutation of n objects taken r at a time

Permutation – Its arrangement

Two element – a b

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

FatherSon

Father is riding

FatherSon

Father is teaching


Three elements a, b, c

Arrangements :

a b c

a c b

b a c

b c a

c a b

c b a

First Second ThirdPlace Place Place


Ist 2nd 3rd

3 ways 2 ways 1 ways

Total ways = 3 + 2 + 1 ? or

3 x 2 x 1 ?


Ist 2nd

2 ways 1 ways

xTotal ways = 2 + 1 = 3 or

2 x 1 = 2


Ist 2nd 3rd

3 ways 2 ways 1 ways

xTotal ways = 3 + 2 + 1 ? or

3 x 2 x 1 ?

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

Mode & Way

Number of style A B?

I

IIBA

Scooter

Cycle

Car

Ways – 2Modes – 4

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

Number of style = 4 x 2 = 8

Independent Process +Dependent Process X

Mode & Way

There are two ways and 4 modes forA B. How many way one can reachB from A?

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

IIIIIIIV

A

BKanpur Lucknow

NewDelhi

Process Dependent

I AI BII AII BIII AIII BIV AIV B

No. of ways = 4 x 2 = 8

Mode & Way

IIIIIIIV

A

BKanpur

LucknowNewDelhi

IV

V

Questions

Illustrative Problem

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

Solution contd..

Odd number

5 ways 3 ways (1/3/5)

Odd numbers = 5 x 3 = 15Total numbers

5 ways 5 ways

Total numbers = 5 x 5 = 25

Even Numbers=10


From the digits 1, 2, 3, 4, 5 how many twodigit numbers can be formed. Whenrepetition is not allowed.

Solution:

5 ways 4 ways

Total = 5 x 4 = 20


There are three questions. Every question canbe 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

TT

F

T

FTF

FT

F

T

FTF


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 = 18 ways

Solution:

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

8ways 10 ways

Girl Boy

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 howmany 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

In a class there are 10 boys and 8 girls.For a quiz competition, a teacher howmany way can select(v) at least one girl while selecting 3 students

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


Principle of Counting

Multiplication Principle : If a job can bedone 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

Questions


Eight children are to be seated on a bench.How many arrangements are possible if theyoungest and eldest child sits at left and rightcorner respectively.

Solution:

We have 6 children to be seated

6 5

Youngest Eldest

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


A class consists of 6 girls and 8 boys. In howmany 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 can’thold more than one position)

Solution :

Girls – 6Boys - 8

Treasurer (Girl)6 ways

Girls – 5Boys - 8

Secretary (Boy)8 ways

Solution contd..

Girls – 5Boys – 7;Total = 12

12 ways

President Vice President

11 waysTotal = 6 x 8 x 12 x 11

Treasurer(Girl)-6 ways Secretary(Boy)-8 ways

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

Questions


1 1 xIf then find x.

9! 10! 11!

Solution :

110 11 121

11! 11!x

9! 10!

11 10 9! 11 10!9! 10!


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

Short cut

2

2

x 3x 2 56

x 3x 54 0

x 9, 6

As x 0 x 6

Permutation

Arrangement of a number of object(s) takensome or all at a fine.

Example : Arrangement of 3 elements out of

5 distinct elements =5.4.3.2.1 5!

5.4.32.1 2!

53

5!P

(5 3)!

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)

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)

?!?!

nr

n!P (r n)

(n r)!

For r = n

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

Questions


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

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 x 2P2

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

= 3!2! = 12

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

= 4P4 – 3P3.2P2

= 4! - 3! 2!= 24 – 12 = 12


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 mins

2P23P3

No. of way = 2P2 x 3P3 = 2! 3! = 24


How many four digits number can beformed 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

Solution

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

3

3

3

3

5

5

5

5

3

3

3

3

53

53

53

53

P

P

P

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.

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 = 5 No. of position available = 3No. of 4 digit number start with ‘3’ = 5P3

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)= 4.5P3 – 5P3 = 3.5P3


In how many way a group photograph of7 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

Solution contd..

Arrangements

Person available – 7Places available – 3

7P4

Total no. of arrangements = 7P3 x 7P4

Person available = 7Places available = 7

7P7arrangements

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 = 3P3 way

Total arrangement = 5.7P4.3P3


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

Solution contd..

Inter subject arrangement

Phy Chem Maths Arrangement = 3P3

Total no. of arrangements = 3P3(3p3 x 2p2 x 4p4)

Class Test

Class Exercise - 1

If are in the

ratio 2 : 1, find the value of n.

n! n!

and2! n 2 ! 4! n 4 !

Solution

Given that

4! n 4 !n!

22! n 2 ! n!

4 3 2! n 4 !

22! n 2 n 3 n 4 !

4 3

2n 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).

Class Exercise - 2

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

Solution

By fundamental principle of counting,the required number = 2 × 2 × 2 = 8(Each place has two choices.)

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?

Solution

(i) Hundred’s place can be filled in 4 ways.

Ten’s place can be filled in 4 ways.

Unit’s place can be filled in 3 ways.

Required number = 4 × 4 × 3 = 48

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

Class Exercise - 4

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

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 least

one digit repeated = 9000 – 4536 = 4464

Class Exercise - 5

There are 5 periods in a school and 6subjects. In how many ways can thetime table be drawn for a day so thatno subject is repeated?

Solution

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

69p 6!

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) 7p5 (d) 35

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)

Class Exercise - 7

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

Solution

Seeta and Geeta can be arranged in 2 ways.

Remaining students can be arranged in 2 ways.

Total ways = 2 × 3 × 2 × 2 = 24

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

Class Exercise - 8

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

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

Class Exercise - 9

There are 5 questions. Each question hastwo options (one answer is correct). Inhow many ways can a student fill up theanswer sheet, when he is asked toattempt all the questions?

Solution

Every question can be answered in two ways.

= 2 ×2 × 2 × 2 × 2 = 25 = 32 ways.

Five questions can be answered in

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?

Solution

(i) Case 1: From left-side, when first student is a boy, then the boys can occupy Ist, 3rd and 5th places. And the girls can occupy 2nd, 4th and 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!

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 girls should 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

Thank you

mathematics. permutation & combination session

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