Match List-I with List-II

List-I	List-II
(A) $^n{C_{r-1}} + ^n{C_r}$	(I) $^n{C_{n-r}}$
(B) $^n{C_r}$	(II) $n + ^1{C_r}$
(C) $^{50}{C_r} = ^{50}{C_{r+2}}$, then r =	(III) 24
(D) $^n{P_3} = 9240$, n=?	(IV) 22

Choose the correct answer from the options given below:

(A) $^nC_{r-1} + ^nC_r$

Using Pascal's Identity (choosing $r$ items from $n+1$ things is the same as choosing $r-1$ from $n$ plus choosing $r$ from $n$):

$^nC_{r-1} + ^nC_r = ^{n+1}C_r$

This matches with (II)

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(B) $^nC_r$

Using the symmetry property of combinations — choosing $r$ items from $n$ is the same as choosing $n - r$ items from $n$:

$^nC_r = ^nC_{n-r}$

This matches with (I)

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(C) $^{50}C_r = ^{50}C_{r+2}$

If $^nC_a = ^nC_b$, then either $a = b$ or $a + b = n$.

Since $r \neq r + 2$, we use the second condition:

$r + (r + 2) = 50$

$2r + 2 = 50$

$2r = 48$

$r = 24$

This matches with (III)

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(D) $^nP_3 = 9240$

$^nP_3 = n(n-1)(n-2)$

We need three consecutive decreasing numbers whose product is $9240$. Taking a rough cube root to estimate the middle value:

$\sqrt[3]{9240} \approx 21$

So we try $n = 22$:

$22 \times 21 \times 20 = 462 \times 20 = 9240$ ✓

Therefore, $n = 22$

This matches with (IV)

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Final matching:

$(A) \rightarrow (II)$

$(B) \rightarrow (I)$

$(C) \rightarrow (III)$

$(D) \rightarrow (IV)$