Match List-I with List-II List-I List-II (A) ^nc_r-1+^nc_r (I) ^nc_n-…

The correct answer is Option (4) → (A)-(II), (B)-(I), (C)-(III), (D)-(IV)

List-I	List-II
(A) ${^nc}_{r-1}+{^nc}_r$	(II) $n+{^1c}_r$
(B) ${^nc}_r$	(I) ${^nc}_{n-r}$
(C) ${^{50}c}_r={^{50}c}_{r+2}$, then r =	(III) 24
(D) ${^np}_3 = 9240$, n=?	(IV) 22

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

This is a standard identity in combinatorics known as Pascal's Rule.

$\begin{pmatrix}{n}\\{r-1} \end{pmatrix}+ \begin{pmatrix}{n}\\{r} \end{pmatrix}= \begin{pmatrix}{n+1}\\{r}\end{pmatrix}$

Therefore, (A) matches with (II).

(B) $^nC_r$

One of the most fundamental properties of combinations is the symmetry property:

$^nC_r = \frac{n!}{r!(n-r)!} = ^nC_{n-r}$

Therefore, (B) matches with (I).

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

In combinations, if $^nC_x = ^nC_y$, then either $x = y$ or $x + y = n$.

• $r = r + 2$ is not possible.
• $r + (r + 2) = 50$

$2r + 2 = 50$

$2r = 48$

$r = 24$

Therefore, (C) matches with (III).

(D) $^nP_3 = 9240$, $n = ?$

The formula for permutations is $^nP_r = \frac{n!}{(n-r)!}$, so $^nP_3 = n(n-1)(n-2)$.

We need to find $n$ such that:

$n(n-1)(n-2) = 9240$

If we test $n = 22$:

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

Therefore, (D) matches with (IV).