Match List-I with List-II

List-I	List-II
(Expression)	(Value)
(A) $\frac{12!}{10!(2!)}$	(I) 110
(B) $^nC_2 = 210$, find n.	(II) 136
(C) $^6P_3 - ^5C_2$	(III) 66
(D) If $^nC_9 = ^nC_8$, find $^nC_{15}$	(IV) 21

Choose the correct answer from the options given below:

(A) $\frac{12!}{10!(2!)}$

This is the combination formula $^{12}C_2$.

Quick shortcut: $^nC_2 = \frac{n(n-1)}{2}$

$^{12}C_2 = \frac{12 \times 11}{2} = \frac{132}{2} = 66$

(A) matches with (III)

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(B) $^nC_2 = 210$, find $n$

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

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

Multiply both sides by $2$:

$n(n-1) = 420$

$n^2 - n - 420 = 0$

Factoring:

$(n-21)(n+20) = 0$

Since $n$ must be positive: $n = 21$

(B) matches with (IV)

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(C) $^6P_3 - ^5C_2$

For $^6P_3$ (arrangement matters, just multiply):

$^6P_3 = 6 \times 5 \times 4 = 120$

For $^5C_2$:

$^5C_2 = \frac{5 \times 4}{2 \times 1} = \frac{20}{2} = 10$

Therefore:

$^6P_3 - ^5C_2 = 120 - 10 = 110$

(C) matches with (I)

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(D) If $^nC_9 = ^nC_8$, find $^nC_{15}$

Using the property: If $^nC_r = ^nC_s$ and $r \neq s$, then $r + s = n$

$9 + 8 = n$

$n = 17$

Now find $^{17}C_{15}$

Using the shortcut $^nC_r = ^nC_{n-r}$:

$^{17}C_{15} = ^{17}C_2 = \frac{17 \times 16}{2} = \frac{272}{2} = 136$

(D) matches with (II)

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Final Matching:

(A) → (III) $= 66$

(B) → (IV) $= 21$

(C) → (I) $= 110$

(D) → (II) $= 136$