Match List-I with List-II

List-I	List-II
(A) $P_3^8-C_3^{10}$	(I) 6
${\mathrm{(B)\; P}}_5^8$	(II) 21
(C) $P_4^n$ = 360, then find n	(III) 216
(D) $C_2^n$ = 210, find n	(IV) 6720

Choose the correct answer from the options given below

The correct option is (A) - (III), (B) - (IV), (C) - (1), (D) - (II).

Let us analyze and solve each item in List-I step-by-step to find its corresponding match in List-II.

Recall the formulas for permutations and combinations:
The number of permutations of n objects taken r at a time is given by:
$$P_r^n=\frac{n!}{(n-r)!}$$
The number of combinations of n objects taken r at a time is given by:
$$C_r^n=\frac{n!}{r!(n-r)!}$$

Step-by-step matching:

(A) Evaluation of $P_3^8-C_3^{10}$:
First, calculate the permutation term:
$$P_3^8=8\times 7\times 6=336$$
Next, calculate the combination term:
$$C_3^{10}=\frac{10\times 9\times 8}{3\times 2\times 1}=120$$
Subtracting the two values:
$$P_3^8-C_3^{10}=336-120=216$$
This matches with (III) 216 in List-II.

(B) Evaluation of $P_5^8$:
Using the permutation formula:
$$P_5^8=8\times 7\times 6\times 5\times 4=6720$$
This matches with (IV) 6720 in List-II.

(C) Finding n for $P_4^n=360$:
We can express the permutation equation as:
$$n(n-1)(n-2)(n-3)=360$$
We factorize 360 into a product of 4 consecutive decreasing integers:
$$360=6\times 5\times 4\times 3$$
Comparing both sides, we get:
$$n=6$$
This matches with (I) 6 in List-II.

(D) Finding n for $C_2^n=210$:
Using the combination formula:
$$\frac{n(n-1)}{2}=210$$
Multiplying by 2:
$$n(n-1)=420$$
We factorize 420 into the product of two consecutive integers:
$$21\times 20=420$$
Comparing both sides, we find:
$$n=21$$
This matches with (II) 21 in List-II.

Summary:
• (A) matches with (III)
• (B) matches with (IV)
• (C) matches with (I)
• (D) matches with (II)