Match List-I with List-II

List-I	List-II
(A) $^{75}P_2 - ^{75}C_2$	(I) 504
(B) $^{5}P_5 - ^{10}C_3$	(II) 6
(C) $^{16}C_{13} - ^{8}C_3$	(III) 2775
(D) $^nP_4 = 360$, then find n	(IV) 0

Choose the correct answer from the options given below:

(A) $^{75}P_2 - ^{75}C_2$

$^{75}P_2 = \frac{75!}{73!}$

$= 75 × 74$

$= 5550$

$^{75}C_2 = \frac{75 × 74}{2 × 1}$

$= \frac{5550}{2}$

$= 2775$

$^{75}P_2 - ^{75}C_2 = 5550 - 2775$

$= 2775$

(A) matches with (III)

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(B) $^{5}P_5 - ^{10}C_3$

$^{5}P_5 = 5!$

$= 5 × 4 × 3 × 2 × 1$

$= 120$

$^{10}C_3 = \frac{10 × 9 × 8}{3 × 2 × 1}$

$= \frac{720}{6}$

$= 120$

$^{5}P_5 - ^{10}C_3 = 120 - 120$

$= 0$

(B) matches with (IV)

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(C) $^{16}C_{13} - ^{8}C_3$

Using $^{16}C_{13} = ^{16}C_{16-13} = ^{16}C_3$:

$^{16}C_3 = \frac{16 × 15 × 14}{3 × 2 × 1}$

$= \frac{3360}{6}$

$= 560$

$^{8}C_3 = \frac{8 × 7 × 6}{3 × 2 × 1}$

$= \frac{336}{6}$

$= 56$

$^{16}C_{13} - ^{8}C_3 = 560 - 56$

$= 504$

(C) matches with (I)

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(D) $^nP_4 = 360$

$n(n-1)(n-2)(n-3) = 360$

For $n = 6$:

$6 × 5 × 4 × 3 = 30 × 12$

$= 360$

Therefore $n = 6$

(D) matches with (II)

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Final Matching:

(A) → (III) = 2775

(B) → (IV) = 0

(C) → (I) = 504

(D) → (II) = 6