Match List-I with List-II

Choose the correct answer from the options given below:

(A)-(III), (B)-(IV), (C)-(I), (D)-(II)

The correct answer is Option (3) → (A)-(III), (B)-(IV), (C)-(I), (D)-(II)

(A) Solve $\frac{1}{6!} + \frac{1}{7!} = \frac{x}{8!}$

• Rewrite the denominators in terms of $6!$:

$\frac{1}{6!} + \frac{1}{7 \cdot 6!} = \frac{x}{8 \cdot 7 \cdot 6!}$

• Factor out $\frac{1}{6!}$ on the left side:

$\frac{1}{6!} \left( 1 + \frac{1}{7} \right) = \frac{x}{56 \cdot 6!}$

• Simplify the terms:

$1 + \frac{1}{7} = \frac{x}{56}$

$\frac{8}{7} = \frac{x}{56}$

• Solve for $x$:

$x = \frac{8}{7} \times 56 = 8 \times 8 = 64$

• Match: (A) $\rightarrow$ (III)

(B) Evaluate $\frac{n!}{(n-r)!}$ for $n=6, r=2$

• Substitute the values into the formula:

$\frac{6!}{(6-2)!} = \frac{6!}{4!}$

• Expand the factorial:

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

• Match: (B) $\rightarrow$ (IV)

(C) If $^nC_9 = ^nC_8$, find $^nC_{17}$

• We know the property: if $^nC_x = ^nC_y$, then either $x = y$ or $x + y = n$.
• Since $9 \neq 8$, we have $9 + 8 = n \Rightarrow n = 17$.
• Now, find $^{17}C_{17}$:

$^{17}C_{17} = 1$

• Match: (C) $\rightarrow$ (I)

(D) Calculate $^6P_3 - ^5P_2$

• Calculate the permutations:
• $^6P_3 = \frac{6!}{(6-3)!} = \frac{6!}{3!} = 6 \times 5 \times 4 = 120$
• $^5P_2 = \frac{5!}{(5-2)!} = \frac{5!}{3!} = 5 \times 4 = 20$
• Subtract the values:

$120 - 20 = 100$

• Match: (D) $\rightarrow$ (II)