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) $n = 9, r = 5$
$$\frac{n!}{(n – r)!} = \frac{9!}{(9 – 5)!} = \frac{9!}{4!}$$
$$\frac{9!}{4!} = \frac{9 \times 8 \times 7 \times 6 \times 5 \times 4!}{4!}$$
$$\frac{9!}{4!} = 9 \times 8 \times 7 \times 6 \times 5$$
$$\frac{9!}{4!} = 72 \times 210 = \mathbf{15120}$$