Match List-I with List-II

List-I	List-II
(Expressions)	(Values)
(A) 1/6! + 1/7! = x/8! Find x	(I) 1
(B) Evaluate: $\frac{n!}{(n-r)!}$, n = 6, r = 2	(II) 100
(C) If ⁿC₉ = ⁿC₈, find ⁿC₁₇.	(III) 64
(D) ⁶P₃ - ⁵P₂	(IV) 30

Choose the correct answer from the options given below:

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

$8! = 8 \times 7!$ and $8! = 56 \times 6!$

Converting $\frac{1}{7!}$ to have denominator $8!$:

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

Converting $\frac{1}{6!}$ to have denominator $8!$:

$\frac{1}{6!} = \frac{56}{8!}$

Adding:

$\frac{1}{6!} + \frac{1}{7!} = \frac{56}{8!} + \frac{8}{8!}$

$= \frac{64}{8!}$

Therefore, $x = 64$ → (A) matches with (III)

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For (B): $\frac{n!}{(n-r)!}$ where $n = 6, r = 2$

This is the permutation formula $^6P_2$:

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

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

$= 6 \times 5$

$= 30$

Therefore, (B) matches with (IV)

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

If $^nC_r = ^nC_s$, then either $r = s$ or $r + s = n$

Since $9 \neq 8$:

$9 + 8 = n$

$n = 17$

Finding $^nC_{17} = ^{17}C_{17}$:

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

Therefore, (C) matches with (I)

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For (D): $^6P_3 - ^5P_3$

Calculate $^6P_3$:

$^6P_3 = \frac{6!}{3!}$

$= 6 \times 5 \times 4$

$= 120$

Calculate $^5P_3$:

$^5P_3 = \frac{5!}{2!}$

$= 5 \times 4 \times 3$

$= 60$

$^6P_3 - ^5P_3 = 120 - 60$

$= 100$

Therefore, (D) matches with (II)

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Final matching:

• (A) → (III) 64
• (B) → (IV) 30
• (C) → (I) 1
• (D) → (II) 100

The correct answer is Option 3.