Rbse Solutions for Class 11 maths Chapter 6 Exercise 6.2 | Factorials

Last Updated on November 24, 2025 by Aman Singh

Get detailed, step-by-step solutions for NCERT Class 11 Maths Chapter 6 Exercise 6.2 . Learn to evaluate and simplify expressions involving factorials (e.g., $8!$, $4! – 3!$). Practice algebraic simplification techniques for fractions of factorials, such as $\frac{8!}{6! 2!}$, by expanding the larger factorials. Solve linear equations involving factorials to find the unknown variable $\mathbf{x}$ (Q.4) and evaluate the $\mathbf{P(n, r)}$ formula $\frac{n!}{(n – r)!}$.

This exercise focuses on the definition and arithmetic of factorials, denoted by $n!$, where $n! = n \times (n-1) \times \dots \times 2 \times 1$.


1. Evaluate Factorials

(i) $8!$

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

$$8! = 56 \times 30 \times 24 = 1680 \times 24 = \mathbf{40320}$$

(ii) $4! – 3!$

We can compute them directly:

$$4! = 4 \times 3 \times 2 \times 1 = 24$$

$$3! = 3 \times 2 \times 1 = 6$$

$$4! – 3! = 24 – 6 = \mathbf{18}$$

(Alternatively, $4! – 3! = 4 \cdot 3! – 3! = 3!(4 – 1) = 3! \cdot 3 = 6 \cdot 3 = 18$)


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

Calculate the Left Hand Side (LHS) and Right Hand Side (RHS) separately.

  • LHS: $3! + 4!$$$3! = 6$$$$4! = 24$$$$LHS = 6 + 24 = 30$$
  • RHS: $7!$$$7! = 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 5040$$

Since $30 \ne 5040$,

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


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

We can expand the larger factorial ($8!$) until the term of the smaller factorial ($6!$) appears, allowing cancellation.

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

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

$$\frac{8!}{6! \times 2!} = \frac{56}{2} = \mathbf{28}$$


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

The common denominator on the left side is the largest factorial, $7!$. Rewrite $\frac{1}{6!}$ as $\frac{7}{7 \cdot 6!} = \frac{7}{7!}$:

$$\frac{7}{7!} + \frac{1}{7!} = \frac{x}{8!}$$

$$\frac{7 + 1}{7!} = \frac{x}{8!}$$

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

Cross-multiply:

$$x \cdot 7! = 8 \cdot 8!$$

Expand $8!$ as $8 \cdot 7!$:

$$x \cdot 7! = 8 \cdot (8 \cdot 7!)$$

Divide both sides by $7!$:

$$x = 8 \times 8$$

$$\mathbf{x = 64}$$


5. Evaluate $\frac{n!}{(n – r)!}$

This formula represents the number of permutations $P(n, r)$.

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

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

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

$$\frac{6!}{4!} = 6 \times 5 = \mathbf{30}$$

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

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

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

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

$$\frac{9!}{4!} = 72 \times 210 = \mathbf{15120}$$

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  • Aman Singh

    Aman Singh | M.Sc. Mathematics, RRBMU University Alwar

    A seasoned Mathematics Educator with 7 years of dedicated experience in the field of education. Specializing in simplifying complex mathematical concepts, Aman has a proven track record of helping students master advanced topics. Holds an M.Sc. in Mathematics from RRBMU University, Alwar. Passionate about leveraging conceptual clarity and effective teaching methodologies to drive student success and achievement.

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    For the past 7 years, Aman Singh has been on a mission to redefine math education. Armed with an M.Sc. in Mathematics from RRBMU University Alwar, Aman brings a deep well of knowledge and seven years of classroom insight to every lesson. Specializing in turning student struggle into genuine mastery, Aman believes math isn't just about numbers—it's about building confidence and problem-solving muscle.