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
a) The words start with P and end with S
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 5C4 = 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 5 |
Part 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
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