MCQ
If $C_0+ C_1 + C_2 + ... + C_n = 256$, then $^{2n}C_2$ is equal to:
  • A
    $56$
  • $120$
  • C
    $28$
  • D
    $91$

Answer

Correct option: B.
$120$
If set $S$ has $n$ elements, then $\text{C (n, k)C n, k}$ is the number of ways of choosing $k$ elements from $S.$
Thus, the number of subsets of $\text{SS}$ of all possible values is given by,
$\text{C}(\text{n},0)+\text{C}(\text{n},1)+\text{C}(\text{n},3)+.....+\text{C}(\text{n},\text{n})=2^\text{n}$
Comparing the given equation with the above equation:
$2^\text{n}=256$
$\Rightarrow 2^\text{n}=2^{8}$
$\Rightarrow \text{n}=8$
$\therefore {^\text{2n}}\text{C}_{\text{2}}={^\text{16}}\text{C}_{\text{2}}$
$\Rightarrow {^\text{16}}\text{C}_{\text{2}}=\frac{16!}{2!4!}=\frac{16\times15}{2}=120$

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