Courses
Courses for Kids
Free study material
Offline Centres
More
Store

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

Last updated date: 21st Sep 2024
Total views: 618.6k
Views today: 7.18k

## NCERT Solutions for Class 11 Maths Chapter 6 Permutations and Combinations

Free PDF download of NCERT Solutions for Class 11 Maths Chapter 6 Exercise 6.2 (Ex 6.2) and all chapter exercises at one place prepared by expert teacher as per NCERT (CBSE) books guidelines. Class 11 Maths Chapter 6 Permutations and Combinations Exercise 6.2 Questions with Solutions to help you to revise complete Syllabus and Score More marks. Register and get all exercise solutions in your emails.

 Class: NCERT Solutions for Class 11 Subject: Class 11 Maths Chapter Name: Chapter 6 - Permutations and Combinations Exercise: Exercise - 6.2 Content-Type: Text, Videos, Images and PDF Format Academic Year: 2024-25 Medium: English and Hindi Available Materials: Chapter WiseExercise Wise Other Materials Important QuestionsRevision Notes
Competitive Exams after 12th Science

## Access NCERT Solutions for Class 11 Maths Chapter 6 – Permutations and Combinations

Exercise 6.2

Refer to pages 3-4 for exercise 6.2 in the PDF

1. Evaluate.

(i) $8!$

Ans: A number's factorial is the function that multiplies it by each natural number below it. The product of the first $n$ natural numbers is $n$ factorial, which is written as $n!$ and is given by, $n! = n \times \left( {n - 1} \right)!$.

Now, the given expression is $8!$.

According to the definition of the factorial,

$8! = 8 \times \left( {8 - 1} \right)!$

$8! = 8 \times 7!$

This is further evaluated as,

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

$8! = 40320$

Therefore, the value of $8!$ is found to be 40320.

(ii) $4! - 3!$

Ans: The product of the first $n$ natural numbers is $n$ factorial, which is written as $n!$ and is given by, $n! = n \times \left( {n - 1} \right)!$.

Now, the given expression is $4! - 3!$.

First, consider and evaluate $4!$.

According to the definition of the factorial,

$4! = 4 \times \left( {4 - 1} \right)!$

$4! = 4 \times 3!$

This is further evaluated as,

$4! = 4 \times 3 \times 2 \times 1$

$4! = 24$

Thus, the value of $4!$ is found to be 24.

Now, consider and evaluate for $3!$.

According to the definition of the factorial,

$3! = 3 \times \left( {3 - 1} \right)!$

$3! = 3 \times 2!$

This is further evaluated as,

$3! = 3 \times 2 \times 1$

$3! = 6$

Thus, the value of $3!$ is found to be 6.

Finally, consider the expression, $4! - 3!$ and evaluate.

$4! - 3! = 24 - 6$

$4! - 3! = 18$

Therefore, the value of the expression on evaluation is found to be 18.

2. Is $4! + 3! = 7!?$

Ans: The product of the first $n$ natural numbers is $n$ factorial, which is written as $n!$ and is given by, $n! = n \times \left( {n - 1} \right)!$.

First, consider and evaluate $4!$.

According to the definition of the factorial,

$4! = 4 \times \left( {4 - 1} \right)!$

$4! = 4 \times 3!$

This is further evaluated as,

$4! = 4 \times 3 \times 2 \times 1$

$4! = 24$

Thus, the value of $4!$ is found to be 24.

Now, consider and evaluate for $3!$.

According to the definition of the factorial,

$3! = 3 \times \left( {3 - 1} \right)!$

$3! = 3 \times 2!$

This is further evaluated as,

$3! = 3 \times 2 \times 1$

$3! = 6$

Thus, the value of $3!$ is found to be 6.

Now, consider the expression, $4! + 3!$ and evaluate.

$4! + 3! = 24 + 6$

$4! + 3! = 30$

Further, consider $7!$.

According to the definition of the factorial,

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

$7! = 5040$

Therefore, it is observed that $4! + 3! \ne 7!$.

3. Compute $\dfrac{{8!}}{{6! \times 2!}}$.

Ans: The product of the first $n$ natural numbers is $n$ factorial, which is written as $n!$ and is given by, $n! = n \times \left( {n - 1} \right)!$.

First consider the expression in the numerator, that is, $8!$.

According to the definition of the factorial,

$8! = 8 \times \left( {8 - 1} \right)!$

$8! = 8 \times 7!$

This is further evaluated as,

$8! = 8 \times 7 \times 6!$

Now, consider the expression in the denominator, that is, $6! \times 2!$.

This expression can also be written in a way that is, $6! \times 2 \times 1$.

Consider the expression that is to be determined and evaluated.

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

$\dfrac{{8!}}{{6! \times 2!}} = \dfrac{{8 \times 7}}{2}$

$\dfrac{{8!}}{{6! \times 2!}} = 28$

Therefore, the value of the expression given on evaluation is found to be 28.

4. $\dfrac{1}{{6!}} + \dfrac{1}{{7!}} = \dfrac{x}{{8!}}$

Ans: The product of the first $n$ natural numbers is $n$ factorial, which is written as $n!$ and is given by, $n! = n \times \left( {n - 1} \right)!$.

The given expression is $\dfrac{1}{{6!}} + \dfrac{1}{{7!}} = \dfrac{x}{{8!}}$.

Consider making the denominator of the left-hand side of the given expression equal.

$\dfrac{1}{{6!}} + \dfrac{1}{{7 \times 6!}} = \dfrac{x}{{8!}}$

Take out the common terms aside,

$\dfrac{1}{{6!}}\left( {1 + \dfrac{1}{7}} \right) = \dfrac{x}{{8!}}$

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

Now, rearrange the above equation and solve for $x$.

$x = \dfrac{{8!}}{{6!}}\left( {\dfrac{8}{7}} \right)$

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

$x = 8 \times 8$

$x = 64$

Therefore, the value of $x$ of the expression on evaluation is found to be 64.

5. Evaluate $\dfrac{{n!}}{{\left( {n - r} \right)!}}$, when

(i) $n = 6,r = 2$

Ans: The expression given is $\dfrac{{n!}}{{\left( {n - r} \right)!}}$.

Substitute 6 for $n$ and 2 for $r$ in the given expression, the expression obtained is,

$\dfrac{{6!}}{{\left( {6 - 2} \right)!}}$

On evaluating the denominator, the expression becomes, $\dfrac{{6!}}{{4!}}$.

Now, consider the expression in the numerator, that is $6!$.

According to the definition of the factorial,

$6! = 6 \times \left( {6 - 1} \right)!$

$6! = 6 \times 5!$

$6! = 6 \times 5 \times 4!$

Further, consider the expression $\dfrac{{6!}}{{4!}}$.

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

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

$\dfrac{{6!}}{{4!}} = 30$

Therefore, the value of the expression $\dfrac{{n!}}{{\left( {n - r} \right)!}}$, when $n = 6,r = 2$ is found to be 30.

(ii) $n = 9,r = 5$

Ans: The expression given is $\dfrac{{n!}}{{\left( {n - r} \right)!}}$.

Substitute 9 for $n$ and 5 for $r$ in the given expression, the expression obtained is,

$\dfrac{{9!}}{{\left( {9 - 5} \right)!}}$

On evaluating the denominator, the expression becomes, $\dfrac{{9!}}{{4!}}$.

Now, consider the expression in the numerator, that is $9!$.

According to the definition of the factorial,

$9! = 9 \times \left( {9 - 1} \right)!$

$9! = 9 \times 8!$

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

Further, consider the expression $\dfrac{{9!}}{{4!}}$.

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

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

$\dfrac{{9!}}{{4!}} = 15120$

Therefore, the value of the expression $\dfrac{{n!}}{{\left( {n - r} \right)!}}$, when $n = 9,r = 5$ is found to be 15120.

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

Opting for the NCERT solutions for Ex 6.2 Class 11 Maths is considered as the best option for the CBSE students when it comes to exam preparation. This chapter consists of many exercises. Out of which we have provided the Exercise 6.2 Class 11 Maths NCERT solutions on this page in PDF format. You can download this solution as per your convenience or you can study it directly from our website/ app online.

Vedantu in-house subject matter experts have solved the problems/ questions from the exercise with the utmost care and by following all the guidelines by CBSE. Class 11 students who are thorough with all the concepts from the Maths textbook and quite well-versed with all the problems from the exercises given in it, then any student can easily score the highest possible marks in the final exam. With the help of this Class 11 Maths Chapter 6 Exercise 6.2 solutions, students can easily understand the pattern of questions that can be asked in the exam from this chapter and also learn the marks weightage of the chapter. So that they can prepare themselves accordingly for the final exam.

Besides these NCERT solutions for Class 11 Maths Chapter 6 Exercise 6.2, there are plenty of exercises in this chapter which contain innumerable questions as well. All these questions are solved/answered by our in-house subject experts as mentioned earlier. Hence all of these are bound to be of superior quality and anyone can refer to these during the time of exam preparation. In order to score the best possible marks in the class, it is really important to understand all the concepts of the textbooks and solve the problems from the exercises given next to it.

Do not delay any more. Download the NCERT solutions for Class 11 Maths Chapter 6 Exercise 6.2 from Vedantu website now for better exam preparation. If you have the Vedantu app in your phone, you can download the same through the app as well. The best part of these solutions is these can be accessed both online and offline as well.

## FAQs on NCERT Solutions for Class 11 Maths Chapter 6: Permutations and Combinations - Exercise 6.2

1. Where can I find NCERT solutions for class 11 maths chapter 6 permutations and combinations exercise 6.2?

Teachers at Vedantu have carefully curated the NCERT solutions for the second exercise of the chapter 6 permutations and combinations. The solutions have been prepared according to the latest CBSE pattern and guidelines along with easy understanding of the concepts. Students can easily access them from the Vedantu website or the app.

2. What does the exercise 6.2 of chapter permutations and combinations based on?

The second exercise of the maths chapter 6 is based on the two concepts of permutations when all the objects are distinct and factorial notations. The NCERT has explained these topics along with solved examples from where students can learn and then solve the questions given in the exercise 6.2.

3. How does NCERT solutions for class 11 Maths chapter 6 permutations and combinations exercise 6.2?

The NCERT solutions for exercise 6.2 helps students in better understanding of maths concepts, enhances their knowledge about the more important topics and gives them an idea as to what type of questions may be asked in the examination regarding these topics. Complete step by step solutions for the complete exercise of 6.2 is available on the Vedantu website from where students can download them in the pdf format.

4. How are permutations and factorial notations related to each other?

The arrangements of objects in a given order or set is known as permutations whereas factorials are the number of possible outcomes of a particular event. The permission and factorials are related in such a way that if permutations are the order of events then the factorial represents the number of times they can be arranged or ordered.

5. What is the symbol of factorial and how do we solve them?

The symbol for factorial is !. Lets understand how to solve a factorial with the following example:

5! = 5 × 4 × 3 × 2 × 1 = 120.

The factorial states that multiply all whole numbers from the chosen number down to the 1.