Solve the following


If C0 + C1 + C2 + ... + Cn = 256, then 2nC2 is equal to

(a) 56

(b) 120

(c) 28

(d) 91


(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:


$\Rightarrow 2^{n}=2^{8}$

$\Rightarrow n=8$

$\therefore{ }^{2 n} C_{2}={ }^{16} C_{2}$

$\Rightarrow{ }^{16} C_{2}=\frac{16 !}{2 ! 14 !}=\frac{16 \times 15}{2}=120$



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