Browse Questions

# If $nC_{12}=nC_6$ then $nC_2$ is equal to

$\begin{array}{1 1}(A)\;72\\(B)\;153\\(C)\;306\\(D)\;2556\end{array}$

Toolbox:
• $C(n,r)=\large\frac{n!}{r!(n-r)!}$
Given that $nC_{12}=nC_6$
$nC{n-12}=nC_6$
$n-12=6$
$n=18$
$nC_2=18C_2=\large\frac{18!}{2!16!}$
$\Rightarrow \large\frac{18\times 17\times 16!}{2\times 1\times 16!}$
$\Rightarrow 153$
Hence (B) is the correct answer.