Match List-I with List-II

List-I	List-II
(A) $^8P_3 - ^{10}C_3$	(I) 6
(B) $^8P_5$	(II) 21
(C) $^nP_4 = 360$, then find n.	(III) 216
(D) $^nC_2 = 210$, find n.	(IV) 6720

Choose the correct answer from the options given below:

The formula for permutation is $^nP_r = \frac{n!}{(n-r)!}$

$^8P_3 = \frac{8!}{(8-3)!}$

$= \frac{8!}{5!}$

$= 8 \times 7 \times 6$

$= 336$

The formula for combination is $^nC_r = \frac{n!}{r!(n-r)!}$

$^{10}C_3 = \frac{10!}{3! \times 7!}$

$= \frac{10 \times 9 \times 8}{3 \times 2 \times 1}$

$= \frac{720}{6}$

$= 120$

Therefore:

$^8P_3 - ^{10}C_3 = 336 - 120$

$= 216$

(A) matches with (III) 216

---

$^8P_5 = \frac{8!}{(8-5)!}$

$= \frac{8!}{3!}$

$= 8 \times 7 \times 6 \times 5 \times 4$

$= 6720$

(B) matches with (IV) 6720

---

$^nP_4 = n(n-1)(n-2)(n-3) = 360$

For $n = 6$:

$6 \times 5 \times 4 \times 3 = 360$

(C) matches with (I) 6

---

$^nC_2 = \frac{n(n-1)}{2} = 210$

$n(n-1) = 420$

For $n = 21$:

$21 \times 20 = 420$

$\frac{420}{2} = 210$

(D) matches with (II) 21

---

Final Matching:

(A) → (III) 216

(B) → (IV) 6720

(C) → (I) 6

(D) → (II) 21