# 11th Class Mental Ability Permutations and Combinations Notes - Permutation & Combination

Notes - Permutation & Combination

Category : 11th Class

# Permutation & Combination

Learning objectives

• Factorial
• Permutation
• Combination

Factorial

The factorial/ symbolized by an exclamation mark (!), is a quantity defined for all integers greater than or equal to 0. Mathematically/ the formula for the factorial is as follows.

If n is an integer greater than or equal to I, then

$n\,\,!=n\left( n-1 \right)\left( n-2 \right)\left( n-3 \right)...(3)(2)(1).$

• Example

$1!=1,\,\,2!=2,\,\,3!=6,\,\,4!=4\cdot 3\cdot 2\cdot 1=24,\,\,5!=5\times 4\times 3\times 2\times 1=120$

$6!=6\times 5\times 4\times 3\times 2\times 1=720,\,\,7!=5040\,\,and\,\,8!=40320\,etc.$

The special case 0! is defined to have value 0! = 1.

Permutation

The different arrangements which can be made by taking some or all of the given things or objects at a time is called Permutation.

• Example:

All permutations (arrangements) made with the letters a, b, c by taking two at a time will be (ab, be, ca, ba, ac, cb).

Number of Permutations: Number of all permutations of n things, taking r at a time is:

$^{n}{{P}_{r}}=\frac{n!}{n-r!}=n(n-1)(n-2)(n-3)...\,\,...(n-r+1).$

Note: This is valid only when repetition is not allowed.

• Permutation of n different things taken r at a time. When repetition is Allowed: $n\times n\times n...\,\,...\,\,...\,\,.\,\,.r$times = ${{n}^{r}}$
• Permutation of n things taking all n things at a time= n!
• Out of n objects ${{n}_{1}}$ are alike one type,${{n}_{2}}$ are alike another type, ${{n}_{3}}$ are alike third type, nr are alike another type such that $({{n}_{1}}+{{n}_{2}}+{{n}_{3}}+.......nr)=n$

Number of permutations of these n things are =$\frac{n!}{{{n}_{1}}!{{n}_{2}}!\,\,...\,\,...\,\,.{{n}_{r}}!}$

Combination

Each of the different selections or groups which are made by taking some or all of a number of things or objects at a time is called combination.

The number of combinations of n dissimilar things taken r at a time is denoted by $^{n}{{C}_{r}}\,\,or\,\,C\,\,(n,\,\,r).$

$^{n}{{C}_{r}}=\frac{n!}{r!(n-r)!}=\frac{n(n-1)(n-2).....(n-r+1)}{1.2.3......r}$

Also $^{n}{{C}_{0}}=1;\,\,\,\,\,\,\,\,\,\,\,{{\,}^{n}}{{C}_{n}}=1;$

Note:    (i) $^{n}{{C}_{r}}{{+}^{n}}{{C}_{r-1}}{{=}^{(n+1)}}{{C}_{r}}$

(ii) $^{n}{{C}_{r}}{{=}^{n}}{{C}_{n-r}}$

Important Formula

• In a group of n-members if each member offers a shake hand to the remaining members then the total number of handshakes = $^{n}{{C}_{2}}=\frac{n(n-1)}{1.2}=\frac{n(n-1)}{2}$
• The number of diagonals in a regular polygon of ‘n’ sides is $\frac{n(n-3)}{2}.$
• From a group of m-men and n-women, if a committee of remembers $(r\le m+n)$is to be formed, then the numbers of ways it can be done is equal to $^{(m+n)}{{C}_{r}}.$
• The number of ways a group of r-boys (men) and s-girls (women) can be made out of m boys (men) and n-girls (women) is equal to ${{(}^{m}}{{C}_{r}}{{\times }^{n}}{{C}_{s}}).$
• From a group of m-boys and n-girls the number of different ways that a committee of remembers can be formed so that the committee will have at least one girl is $^{(m+n)}{{C}_{r}}{{-}^{m}}{{C}_{r}}.$

1. If a die is cast and then a coin is tossed, find the number of all possible outcomes.

(a) 11                                                    (b) 12

(c) 10                                                    (d) 15

(e) None of these

Ans.     (b)

Explanation: A die can fall in 6 different ways showing six different points 1, 2, 3, 4, 5, 6, ... and a coin can fall in 2 different ways showing head (H) or tail (T).

$\therefore$ The number of all possible outcomes from a die and a coin $=6\times 2=12.$

1. There are 6 trains running between Indore and Bhopal. In haw many ways can a go from Indore to Bhopal and return by a different train?

(a) 25                                                    (b) 35

(c) 30                                                    (d) 20

(e) None of these

Ans.     (c)

Explanation: A man can go from Indore to Bhopal in 6 ways by any one of the 6 trains available. Then he can return from Bhopal to Indore in 5 ways by the remaining 5 trains, since he cannot return by the same train by which he goes to Bhopal from Indore.

Thus, the required number of ways $=6\times 5=30.$

1. Find the value of $^{\mathbf{9}}{{\mathbf{P}}_{\mathbf{3}}}\mathbf{.}$

(a) 504                                                  (b) 309

(c) 405                                                  (d) 600

(e) None of these

Ans.     (a)

Explanation:

$^{9}{{P}_{3}}=\frac{9!}{6!}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( \,\,{{\therefore }^{n}}{{P}_{r}}=\frac{n!}{(n-r)!} \right)$.

$=\frac{9\times 8\times 7\times 6!}{6!}=9\times 8\times 7=504$

1. In how many different ways can the letters of the word 'STRESS' be arranged?

(a) 120                                                  (b) 420

(c) 840                                                  (d) 240

(e) None of these

Ans.     (a)

Explanation: The word 'STRESS' has a total of six letters (n = 6) out of which a group of three letters are same (a = 3)

$\therefore$ The letters can be arranged in $\frac{n!}{a!}=\frac{6!}{3!}=\frac{720}{6}=120.$

1. Find the value of $^{\mathbf{8}}{{\mathbf{C}}_{\mathbf{3}}}\mathbf{.}$

(a) 56                                                    (b) 8!

(c) 65                                                    (d) ${{3}^{8}}$

(e) None of these

Ans.     (a)

Explanation: $^{8}{{C}_{3}}=\frac{8!}{3!\,\,\cdot \,\,5!}=\frac{8\times 7\times 6}{6}=56$

Directions: Study the given information carefully and answer the questions that follow:

A committee of five members is to be formed out of 3 trainees, 4 professors and 6 research associates. In how many different ways can this be done if-

1. The committee should have all 4 professors and 1 research associate or all 3 trainees and 2 professors?

(a) 13                                                    (b) 12

(c) 24                                                    (d) 35

(e) None of these

Ans.     (b)

Explanation: Five member team with 4 professors and 1 research associate can be selected in $^{4}{{C}_{4}}{{\times }^{6}}{{C}_{1}}=1\times 6=6$ ways. Five member team with 3 trainees and 2 professors can be selected in $^{3}{{C}_{3}}{{\times }^{4}}{{C}_{2}}=1\times 6=6$ways.

$\therefore$ Total number of ways of selecting the committee $=6+6=12.$

1. The committee should have 2 trainees and 3 research associates?

(a) 15                                                    (b) 45

(c) 60                                                    (d) 75

(e) None of these

Ans.     (c)

Explanation: 2 trainees and 3 research associates can be selected in$^{3}{{C}_{2}}{{\times }^{6}}{{C}_{3}}=3\times 20=60$ways.

#### Other Topics

##### Notes - Permutation and Combination

You need to login to perform this action.
You will be redirected in 3 sec