NCERT Solutions for Exercise 7.2 Class 11 Maths Chapter 7

NCERT Solutions for Exercise 7.2 Class 11 Maths Chapter 7 - Permutations and Combinations

Edited By Ravindra Pindel | Updated on Jul 12, 2022 06:45 PM IST

In the previous exercise, you have already learned about the fundamental principle of counting. In this article, you will get NCERT solutions for Class 11 Maths Chapter 7 exercise 7.2 where you will learn about the counting technique called "Permutations". It is a very important concept based on the fundamental principle of counting. Also, you will get questions from topics such as permutations when all the objects are distinct and factorial notation in the Class 11 Maths chapter 7 exercise 7.2 solutions. You should try to solve all the problems given in the Class 11 Maths chapter 7 exercise 7.2. You can take help from Class 11 Maths chapter 7 exercise 7.2 solutions if you are finding difficulty while these NCERT problems.

JEE Main Scholarship Test Kit (Class 11): Narayana | Physics Wallah | Aakash | Unacademy

Suggested: JEE Main: high scoring chapters | Past 10 year's papers

This Story also Contains

Permutation And Combinations Class 11 Chapter 7 Exercise 7.2
Question:1(i) Evaluate
More About NCERT Solutions for Class 11 Maths Chapter 7 Exercise 7.2:-
Benefits of NCERT Solutions for Class 11 Maths Chapter 7 Exercise 7.2:-
NCERT Solutions of Class 11 Subject Wise
Subject Wise NCERT Exampler Solutions

NCERT solutions for Class 11 Maths Chapter 7 exercise 7.2 are not only important to understand the permutations but are very useful in the chapter like probability also. Also, check NCERT solutions if you are looking for the NCERT solutions from Classes 6 to 12 for Science and Maths at one place.

Also, see

Permutation And Combinations Class 11 Chapter 7 Exercise 7.2

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 $n=9,r=5$

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

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

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

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

Benefits of NCERT Solutions for Class 11 Maths Chapter 7 Exercise 7.2:-

Class 11 Maths chapter 7 exercise 7.2 solutions are designed in a descriptive manner which could be understood by an average student also.
Class 11 Maths chapter 7 exercise 7.2 solutions are created based on the CBSE guideline, so you rely upon these solutions
Class 11 Maths chapter 7 execise 7.2 are useful for the quick revision before the CBSE exam also.

Also see-

JEE Main Highest Scoring Chapters & Topics

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NCERT Solutions of Class 11 Subject Wise

Subject Wise NCERT Exampler Solutions

Happy learning!!!

Frequently Asked Questions (FAQs)

1. Evaluate 3!

3! = 1x2x3 = 6

2. Evaluate 6 ! – 4!

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

3. Is 2 ! + 4 ! = 6 ! ?

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

6! = 1x2x3x4x5x6 = 720

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

4. 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.

5. What is the wieghtage of permutations & combinations in the JEE Main exam ?

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

6. What is the wieghtage of sequences & series in the JEE Main exam ?

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

Also Read

NCERT Class 11 Chemistry Chapter 4 Notes Chemical Bonding and Molecular Structure - Download PDF

Hydrogen Class 11th Notes - Free NCERT Class 11 Chemistry Chapter 9 Notes - Download PDF

The S-Block Elements Class 11th Notes - Free NCERT Class 11 Chemistry Chapter 10 Notes - Download PDF

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NCERT Class 11 Physics Chapter 3 Notes Motion in a straight Line - Download PDF

NCERT Solutions for Exercise 10.1 Class 11 Maths Chapter 10 - Straight Lines

NCERT Solutions for Miscellaneous Exercise Chapter 16 Class 11 - Probability

NCERT Exemplar Class 11 Maths Solutions Chapter 12 Introduction to Three Dimensional Geometry

NCERT Exemplar Class 11 Maths Solutions Chapter 10 Straight Lines

NCERT Exemplar Class 11 Maths Solutions Chapter 6 Linear Inequalities

NCERT Exemplar Class 11 Maths Solutions Chapter 16 Probability

NCERT Exemplar Class 11 Maths Solutions Chapter 14 Mathematical Reasoning

NCERT Exemplar Class 11 Maths Solutions Chapter 7 Permutations and Combinations

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