Match List-I with List-II

Choose the correct answer from the options given below:

(A)-(III), (B)-(IV), (C)-(I), (D)-(II)

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

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

Using the definition of factorials:

$\frac{12!}{10! \cdot 2!} = \frac{12 \times 11 \times 10!}{10! \times 2 \times 1} = \frac{132}{2} = 66$

Match: (A) $\rightarrow$ (III)

(B) Finding $n$ for ${}^nC_2 = 210$

The formula for combinations is ${}^nC_r = \frac{n!}{r!(n-r)!}$. For $r=2$:

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

$n(n-1) = 420$

Since $21 \times 20 = 420$, we find that $n = 21$.

Match: (B) $\rightarrow$ (IV)

(C) Calculation of ${}^6P_3 - {}^5C_2$

• Permutation ${}^6P_3 = 6 \times 5 \times 4 = 120$
• Combination ${}^5C_2 = \frac{5 \times 4}{2 \times 1} = 10$

Calculation: $120 - 10 = 110$.

Match: (C) $\rightarrow$ (I)

(D) Finding ${}^nC_{15}$ given ${}^nC_9 = {}^nC_8$

Using the property ${}^nC_x = {}^nC_y \Rightarrow x+y = n$:

$n = 9 + 8 = 17$

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

${}^{17}C_{15} = {}^{17}C_{17-15} = {}^{17}C_2 = \frac{17 \times 16}{2} = 17 \times 8 = 136$

Match: (D) $\rightarrow$ (II)