NCERT Solutions for Class 11 Maths Chapter 6 Exercise 6.2 - Permutations and Combinations

Komal MiglaniUpdated on 06 May 2025, 02:25 PM IST

In our daily lives we often need to assess the possible number of arrangements in various situations related to the seating arrangement of people, arranging digits or making an arrangement of letters and numbers. When the sample size is small, it is easy to arrange by the conventional method, but when the sample is of large numbers, we need better advanced accounting techniques, which we study under the topic named permutations and combinations.

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NCERT Solutions Class 11 Maths Chapter 6 Exercise 6.2
Topics covered in Chapter 6 Permutations and Combinations Exercise 6.2
NCERT Solutions of Class 11 Subject Wise
Subject-Wise NCERT Exemplar Solutions

Exercise 6.2 of Class 11 Maths in NCERT focuses on topics related to permutations when all the objects are distinct, and the factorial method is used to understand the count and organise the objects in different orders. These concepts further help in other topics such as probability. The NCERT Solutions discuss the step-by-step methodology to apply these techniques effectively.

NCERT Solutions Class 11 Maths Chapter 6 Exercise 6.2

Question:1(i) Evaluate $\; 8 !$

Answer:

Factorial can be given as :

$8!=8\times 7\times 6\times 5\times \times 4\times 3\times 2\times 1=40320$

Question:1(ii) Evaluate $\; 4!-3!$

Answer:

Factorial can be given as :

$(ii)\; 4!-3!$

$=( 4\times 3\times 2\times 1)-(3\times 2\times 1)$

$=24-6$

$=18$

Question:2 Is $3!+4!=7!\; ?$

Answer:

Factorial can be given as :

To prove : $3!+4!=7!$

R.H.S : $3!+4!$

$=(3\times 2\times 1)+( 4\times 3\times 2\times 1)$

$=6+24$

$=30$

L.H.S : $7!$

$=7\times 6\times 5\times 4\times 3\times 2\times 1=5040$

$L.H.S\neq R.H.S$

Question:3 Compute $\frac{8!}{6!\times 2!}$

Answer:

To compute the factorial :

$\frac{8!}{6!\times 2!}=\frac{8\times 7\times 6!}{6!\times 2\times 1}$

$=\frac{8\times 7}{ 2}$

$=4\times 7=28$

So the answer is 28

Question:4 If $\frac{1}{6!}+\frac{1}{7!}=\frac{x}{8!},$ find x

Answer:

Factorial can be given as :

To find x : $\frac{1}{6!}+\frac{1}{7!}=\frac{x}{8!},$

$R.H.S :\, \, \frac{1}{6!}+\frac{1}{7!}$

$= \frac{1}{6!}+\frac{1}{7\times 6!}$

$= \frac{1}{6!}(1+\frac{1}{7})$

$= \frac{1}{6!}(\frac{8}{7})$

$L.H.S:\, \, \frac{x}{8!},$

$=\frac{x}{8\times 7\times 6!}$

Given : $L.H.S= R.H.S$

$\frac{1}{6!}\left ( \frac{8}{7} \right )=\frac{x}{8\times 7\times 6!}$

$\Rightarrow \, \, 8=\frac{x}{8}$

$\Rightarrow \, \, x=8\times 8=64$

Question:5(i) Evaluate $\frac{n!}{(n-r)!},$ when

$\; n=6,r=2$

Answer:

To evaluate $\frac{n!}{(n-r)!},$

Put $n=6,r=2$

$\frac{n!}{(n-r)!}=\frac{6!}{(6-2)!}$

$=\frac{6!}{4!}$

$=\frac{6\times 5\times 4!}{4!}$

$=6\times 5=30$

Question:5(ii) Evaluate $\frac{n!}{(n-r)!},$ when

$\; n=9,r=5$

Answer:

To evaluate $\frac{n!}{(n-r)!},$

Put $n=9,r=5$

$\frac{n!}{(n-r)!}=\frac{9!}{(9-5)!}$

$=\frac{9!}{4!}$

$=\frac{9\times 8\times 7\times 6\times 5\times 4!}{4!}$

$=9\times 8\times 7\times 6\times 5=15120$

Also, see

Topics covered in Chapter 6 Permutations and Combinations Exercise 6.2

1. Permutations:
Permutations mean the different ways of arranging a set of objects in a specific order. In permutations, the order of arrangement is important. Understanding of this concept further helps in solving complex problems.

2. Permutations when all the objects are distinct:
When all the objects in a group are different from each other, the number of ways they can be arranged depends on how many objects are taken at a time. Each arrangement gives a unique permutation.

3. Factorial Notation:
Factorial notation is a shorthand used in counting arrangements. It is represented by an exclamation mark (!) and means the product of all natural numbers from 1 up to that number (e.g., 4! = 4 × 3 × 2 × 1 = 24).

Also Read

NCERT Solutions of Class 11 Subject Wise

Follow the links to get your hands on subject-wise NCERT textbook solutions to ace your exam preparation.

NCERT Solutions for Class 11 Maths

NCERT Solutions for Class 11 Physics

NCERT Solutions for Class 11 Chemistry

NCERT Solutions for Class 11 Biology

Subject-Wise NCERT Exemplar Solutions

Follow the links provided below for various subjects to practise exemplar questions for the examination.

NCERT Exemplar Solutions for Class 11 Maths

NCERT Exemplar Solutions for Class 11 Physics

NCERT Exemplar Solutions for Class 11 Chemistry

NCERT Exemplar Solutions for Class 11 Biology

Frequently Asked Questions (FAQs)

Q: Evaluate 3!

3! = 1x2x3 = 6

Q: Evaluate 6 ! – 4!

6! -4! = 1x2x3x4x5x6 - 1x2x3x4 = 720 - 24 = 696

Q: Is 2 ! + 4 ! = 6 ! ?

2 ! + 4 ! = 1x2 + 1x2x3x4 = 2 + 24 = 26

6! = 1x2x3x4x5x6 = 720

Hence 2 ! + 4 ! is not equal to 6 !.

Q: What is the wieghtage of algebra in the CBSE Class 11 Maths ?

Algebra has 30 marks weightage which is the highest in the CBSE Class 11 Maths final exam.

Q: What is the wieghtage of permutations & combinations in the JEE Main exam ?

Permutations & Combinations has 3.3 % weightage in the maths JEE Main.

Q: What is the wieghtage of sequences & series in the JEE Main exam ?

Sequences & Series has 6.6 % weightage in the maths JEE Main.

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