Question:
If C0 + C1 + C2 + ... + Cn = 256, then 2nC2 is equal to
(a) 56
(b) 120
(c) 28
(d) 91
Solution:
(b) 120
If set $S$ has $n$ elements, then $C(n, k)$ is the number of ways of choosing $k$ elements from $S$.
Thus, the number of subsets of $S$ of all possible values is given by
$C(n, 0)+C(n, 1)+C(n, 3)+\ldots+C(n, n)=2^{n}$
Comparing the given equation with the above equation:
$2^{n}=256$
$\Rightarrow 2^{n}=2^{8}$
$\Rightarrow n=8$
$\therefore{ }^{2 n} C_{2}={ }^{16} C_{2}$
$\Rightarrow{ }^{16} C_{2}=\frac{16 !}{2 ! 14 !}=\frac{16 \times 15}{2}=120$