NCERT Solutions for Class 11 Maths Chapter 6 Permutations And Combinations Ex 6.2

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 Solution Class 11 Maths of Chapter 1 All Exercises

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.

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