Question

If $C_0+C_1+C_2+...+C_n=256$ = then ${}^{2n}{C}_2$ is equal to

• A. 56
• B. 120
• C. 28
• D. 91

Answer 1

The correct option is 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

$C(n,0)+C(n,1)+C(n,3)+...+C(n,n)=2^n$

Comparing the given equation with the above equation :

$2^n=256$

$\Rightarrow 2^n=2^8$

$\Rightarrow n=8$

${\therefore }^{2n}{C}_2{=}^{16}{C}_2$

${\Rightarrow }^{16}{C}_2=\frac{16!}{2!14!}=\frac{16\times 15}{2}=120$