Match List-I with List-II List-I List-II (A) ^8P_3-^10C_3 (I) 6 (B) ^…

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

List-I	List-II
(A) ${^8P}_3-{^{10}C}_3$	(III) 216
(B) ${^8P}_5$	(IV) 6720
(C) ${^nP}_4=360$, then find $n$.	(I) 6
(D) ${^nC}_2 = 210$, find $n$.	(II) 21

(A) $^8P_3 - ^{10}C_3 \rightarrow$ (III) 216

• Calculate $^8P_3$: $8 \times 7 \times 6 = 336$
• Calculate $^{10}C_3$: $\frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 10 \times 3 \times 4 = 120$
• Subtract: $336 - 120 = \mathbf{216}$

(B) $^8P_5 \rightarrow$ (IV) 6720

• Calculation: $8 \times 7 \times 6 \times 5 \times 4 = \mathbf{6720}$

(C) $^nP_4 = 360 \rightarrow$ (I) 6

• The expression for $^nP_4$ is $n(n-1)(n-2)(n-3) = 360$.
• We look for four consecutive integers whose product is 360.
• Testing $n = 6$: $6 \times 5 \times 4 \times 3 = 360$.
• Therefore, $n = 6$.

(D) $^nC_2 = 210 \rightarrow$ (II) 21

• The expression for $^nC_2$ is $\frac{n(n-1)}{2} = 210$.
• Multiply by 2: $n(n-1) = 420$.
• Since $21 \times 20 = 420$, we find that $n = 21$.