# Permutations and Combinations Class 11 Chapter 7

## Permutations and Combinations ### Fundamental principle of counting

If an event can occur in m different ways, following which another event can occur in n different ways, then the total number of occurrence of the events in the given order is m×n.

1. How many three digit numbers can be formed using the digits 1,2,3,4,5 if

a) repetition of the digits is allowed

b) repetition of the digits is not allowed

a) 5×5×5 = 125

b) 5×4×3 = 60

2. How many three digit numbers can be formed using the digits

1,2,3,4,5,6 if

a) repetition of the digits is allowed

b) repetition of the digits is not allowed

a) 6  ×6 ×3 =108

b) 4 ×5 ×3= 60

3. How many 5 digit telephone numbers can be constructed if each number starts with 67 and no digit is repeated

8 ×7 ×6 = 336

4.How many 4 digit numbers are there with no digit repeated

9 ×9 ×8 ×7 = 4536

Factorial

The continued product of first n natural numbers is called n factorial and is denoted by n!

n! = 1 ×2 ×3 ×….(n-1) ×n

0! =1

1! =1

2! = 1 ×2 =2

3! = 1 ×2 ×3 = 6

4! =1 ×2 ×3 ×4 = 24

5! = 1 ×2 ×3 ×4 ×5

= 5 ×4 ×3!

10! = 10 ×9!

= 10 ×9 ×8!

n! =n(n-1)!

= n(n-1)(n-2)!

1.     Find 4! + 3!

= 24+6 = 30

2. Find  = 90

3.  4.     (n+1)! = 90(n-1)! Find n

(n+1)n(n-1)! = 90(n-1)!

n(n+1) = 90

n(n+1) =9×10

n=9

5. =  Find x   ### Permutations

Consider the letters A,B,C

Arrangements taking 2 at a time

AB,BA,AC,CA,BC,CB

Arrangements taking 3 at a time

ABC, ACB, BAC, BCA, CAB, CBA

Permutation is used to find the number of arrangements.

nPr is the total number of arrangements out of n objects taking r at a time without repetition.

nPr = nP1 = n

nP2= n(n-1)

nP3 = n(n-1)(n-2)

nPn = n!

10P1 = 10

10P2 = 10×9

10P3 = 10×9×8

Number of arrangements of n objects taking all at a time = n!

The number of arrangements of 5 boys on a bench = 5! = 120

The number of arrangements of 10 girls on a line for a photograph

= 10!

Q. In how many ways can 7 books be arranged of a shelf so that 3 particular books will come together

Consider the particular 3 books as one unit. This unit can be arranged in 3! Ways. Remaining 4 books and this unit (total 5) can be arranged in 5! ways

Total = 3! × 5! = 6 × 120 = 720

Q. How many words can be formed using the letters of the word MONDAY using each letter once if

(a) 4 letters are used at a time

(b) All  letters are used at a time

(a) 6P4 = 6 ×5 ×4 ×3 = 360

(b) 6!  = 720

Q. Consider the word DAUGHTER. The letters of the word are rearranged.

(a) find the total number of words

8! = 40320

(b) Find the number of words starts with U

7! = 5040

(C) = Find the number of words starts with G and end with R

6! = 720

(d)Find the number of words in which vowels are together

3! × 6! = 6 × 720 = 4320

(e) Find the number of words in which vowels are not together

= 40320 – 4320 = 36000

Q. Find the number of arrangements of the letters of the word BANANA

A-3, N- 2

Total = Q. In how many ways can the letters of the word PERMUTATIONS be arranged if b) Vowels are together

5! × c) There are always 4 letters between P and S

The letters except P and S can be arranged in ways.  P and S can be arranged in 14 ways

Total = ×14

Q. =   find n  Q. 5Pr = 6Pr-1 find r (7-r)(6-r) = 6

42-7r-6r+r2 = 6

r2-13r+36 =0

(r-9)(r-4) = 0

r = 9,4

r = 4

### Combination

Combination is a mathematical method to find number of selections.

nCr is the total number of selection out of n objects taking r at a time(r≤n). nCr = nCn-r

10C7 = 10C10-7 = 10C3

100C98 = 100C100-98 = 100C2

Q. nC8 = nC9. Find nC2

n = 8+9 = 17.

nC2 = 17C2

= =136

nCr  r! = nPr

Q. How many chords can be drawn through 21 points on a circle

A chord is formed by joining any two points on a circle. To form a chord, we need 2 points. There are 21 points

Number of chords = 21C2  = =210

Q. Find the number of diagonals of an octagon

Number of diagonals of a polygon = nC2 – n

Number of diagonals of an octagon = 8C2 – 8 = 28 – 8 = 20

Q.  In how many ways can 2 boys and 3 girls be selected from 4 boys and 5 girls Q. In how many ways can a cricket team of 11 be selected from 17 in which 5 players can bowl and each team of 11 must contain 4 bowlers

12 batsmen 5 bowlers

7 batsmen  4 bowlers

12C7   5C= 3960

Q. The English alphabet has 5 vowels and 21 consonants. How many words with 2 different vowels and 2 different consonants can be formed?

Total number of selections = 5C2 X 21C2 = 2100

Using each selection of 4 letters, 4! Words can be formed

Total number of words = 2100 X 4! = 5040

Q. How many words with or without meaning each of 2 vowels and 3 consonants can be formed from the letters of the word DAUGHTER

3 vowels 5 consonants

2 vowels 3 consonants

Total number of selections = 3C2  5C3 = 30

Using each selection of 5 letters, 5! Words can be formed

Total words = 30  5! = 3600

Q.  In an examination, a question paper consists of 12 questions divided into two parts part I and part II containing 5 and 7 questions respectively. A student is required to attempt 8 questions in all selecting atleast 3 from each part. In how many ways can a student select the questions?

 Part I 5 Part II 7 3 5 4 4 5 3

Total = 5C3.7C5 + 5C4.7C4 + 5C5.7C3 = 420.

Q. 2nC3: nC3 = 12 : 1 find n 2n-1 = 3(n-2)

2n-1= 3n-6

3n-2n = -1+6

n= 5

Q. In a school there are 20 teachers. In how many ways can a principal and vice principal can be selected

The principal can be selected in 20 ways and vice principal can be selected in 19 ways.

Total = 20×19 = 380

Q. In a meeting, everyone shake hands with each other. If 45 shake hands were exchanged, find the number of people attended the meeting

Let n be the number of people attended the meeting

Number of shake hands = nC2

nC2 = 45 n(n-1) = 90 = 10×9

n=10

Q. If nPr = 840, nCr = 35, find r

nCr × r! = nPr

35 × r! = 840

r! = 840/35= 24

r = 4

Q.  22Cr+2 = 22C2r-1   find r

r+2 = 2r-1

2r-r = 2+1

r=3

r+2+2r-1 = 22

3r+1 = 22

3r = 22-1 = 21

r = 21/3= 7

Q.  A committee of 7 has to be formed from 9 boys and 4 girls. In how many

ways can this be done when the committee consists of

a) exactly 3 girls

9 boys       4 girls

4 3 9C4x4C3= 504

b) atleast 3 girls

9 boys       4 girls

4 3 9C4x4C3= 504

3 4 9C3 x4C4=84

Total =504 + 84 = 588

c) atmost 3 girls

9 boys       4 girls

4 3 9C4 x4C3 = 504

5 2 9C5 x4C2 =756

6 1 9C6 x4C1 =336

7 0 9C7 x4C0 =36

Total = 504 + 756 + 336 + 36 = 1632