# If $nC_8=nC_2$ Find $nC_2$

$\begin{array}{1 1}(A)\;40\\(B)\;45\\(C)\;50\\(D)\;55\end{array}$

We have $nC_r=nC_{n-r}$
$nC_2=nC_{n-2}$
$nC_8=nC_{n-2}$
$n-2=8$
$n=8+2$
$n=10$
$nC_2=10C_2$
$\Rightarrow \large\frac{10\times 9}{1\times 2}$
$\Rightarrow 45$
$nC_2=45$
Hence (B) is the correct answer.