Permutations And Combinations Exercise 6.2 NCERT Solutions Class 11 Math

Anand Classes presents detailed NCERT Solutions for Class 11 Maths Chapter 6 Permutations and Combinations Exercise 6.2, helping students grasp the concepts of permutations, arrangements, and counting principles with accuracy. These step-by-step solutions follow the latest CBSE and NCERT guidelines, ensuring that every question is solved with proper reasoning and mathematical clarity. Students can easily practice different types of permutation problems and strengthen their understanding of key formulas and examples. Click the print button to download study material and notes.

NCERT Question 1. Evaluate
(i) $8!$ (ii) $4! – 3!$

Solution:

(i) $8!$

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

$$
= 56 \times 30 \times 12 \times 2
$$

$$
= 1680 \times 24
$$

$$
= 40320
$$

Hence,
$$\boxed{8! = 40320}$$

(ii) $4! – 3!$

$$
4! – 3! = 4 \times 3! – 3!
$$

$$
= 3! \times (4 – 1)
$$

$$
= 3! \times 3
$$

$$
= 6 \times 3 = 18
$$

Hence,
$$\boxed{4! – 3! = 18}$$

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

Solution:

L.H.S:
$$
3! + 4! = 3! + 4 \times 3! = 3!(1 + 4) = 3! \times 5 = 6 \times 5 = 30
$$

R.H.S:
$$
7! = 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 5040
$$

Since $30 \ne 5040$,

$$\boxed{3! + 4! \ne 7!}$$

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

Solution:

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

Cancelling $6!$ from numerator and denominator:

$$
= \dfrac{8 \times 7}{2 \times 1} = 4 \times 7 = 28
$$

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

NCERT Question 4. If $\dfrac{1}{6!} + \dfrac{1}{7!} = \dfrac{y}{8!}$, find $y$.

Solution:

L.C.M of $6!$ and $7!$ is $7!$

So,
$$
\dfrac{1}{6!} + \dfrac{1}{7!} = \dfrac{7 + 1}{7!} = \dfrac{8}{7!}
$$

Equating with $\dfrac{y}{8!}$, we get

$$
\dfrac{8}{7!} = \dfrac{y}{8!}
$$

$$
\Rightarrow y = 8 \times 8 = 64
$$

Hence,
$$\boxed{y = 64}$$

Question 5. Evaluate $\dfrac{n!}{(n – r)!}$
(i) When $n = 6, r = 2$ (ii) When $n = 9, r = 5$

Solution:

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

$$
\dfrac{n!}{(n – r)!} = \dfrac{6!}{(6 – 2)!} = \dfrac{6!}{4!}
$$

$$
= \dfrac{6 \times 5 \times 4!}{4!} = 6 \times 5 = 30
$$

Hence,
$$\boxed{30}$$

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

$$
\dfrac{n!}{(n – r)!} = \dfrac{9!}{(9 – 5)!} = \dfrac{9!}{4!}
$$

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

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

$$
= 72 \times 7 \times 30 = 72 \times 210 = 15120
$$

Hence,
$$\boxed{15120}$$

Master factorial problems, permutations, and combinations easily with detailed NCERT solutions and step-by-step explanations by Anand Classes for Class 11 Maths, JEE Main, NDA, and CUET preparation.