Question

If ${}^n{C}_8=$ ${}^n{C}_2$, find ${}^n{C}_2$.

Answer 1

We are given, $\left(\frac{n}{8}\right)=\left(\frac{n}{2}\right)$

It is only possible if $n=8+2$.

So, $n=10$

So, $\left(\frac{n}{2}\right)=\left(\frac{10}{2}\right)=\frac{10!}{2!(10-2)!}$

$=\frac{10!}{2!\times 8!}$

$=\frac{10\times 9\times 8!}{2\times 8!}$

$\left(\frac{n}{2}\right)=45$