Question

Prove that: ${}^{2n}{C}_n=\frac{2^n\{1\cdot 3\cdot 5\cdot ....(2n-1\left)\right\}}{n!}$

Answer 1

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

$=\frac{2n(2n-1)(2n-2)(2n-3)(2n-4)..........3.2.1}{n!n!}$

$=\frac{2^n\left\{n\right(2n-1\left)\right(n-1\left)\right(2n-3\left)\right(n-2).........3.1.1\}}{n!n!}$

$=\frac{2^{n}n!\left\{\right(2n-1\left)\right(2n-3).........3.1\}}{n!n!}$

$=\frac{2^n\left\{\right(2n-1\left)\right(2n-3).........3.1\}}{n!}$