Question

If $\left(\begin{array}{c}n\\ 8\end{array}\right)=\left(\begin{array}{c}n\\ 2\end{array}\right)$ then find $\left(\begin{array}{c}2n\\ n\end{array}\right)$

Answer 1

${}^n{C}_8{=}^n{C}_2$

$\frac{n!}{(n-8)!8!}=\frac{n!}{(n-2)!2!}$

$(n-2)!2!=(n-8)!8!$

$2!\times (n-2)(n-3)(n-4)(n-5)(n-6)(n-7)(n-8)!=(n-8)!8\times 7\times 6\times 5\times 4\times 3\times 2!$

$2!\times (n-2)(n-3)(n-4)(n-5)(n-6)(n-7)=8\times 7\times 6\times 5\times 4\times 3$

To find ${}^{2n}{C}_n=\frac{\left(2n\right)!}{(2n-n)!n!}$

$=\frac{\left(2n\right)!}{n!n!}\Rightarrow \frac{\left(2n\right)(2n-1)(2n-2)!}{n(n-1)!n!}$

$\Rightarrow \frac{4(2n-1)}{n!}$