Question

Prove that, $\frac{{}^{4n}{C}_{2n}}{{}^{2n}{C}_n}=\frac{1\cdot 3\cdot 5....(4n-1)}{[1\cdot 3\cdot 5....(2n-1){]}^2}$

Answer 1

$\frac{2n!}{n!}=\frac{\left(2n\right)(2n-1)(2n-3)\cdots 5\cdot 3\cdot 1}{n!}=\frac{2^n\left[n\right(n-1\left)\right(n-2)\cdots 2.1][1\cdot 3\cdot \mathrm{5....}(2n-1\left)\right]}{n!}=\frac{2^n(n!)1\cdot 3\cdot \mathrm{5....}(2n-1)}{n!}$

$=2^n(1\cdot 3\cdot 5....(2n-1))$

Similarly $\frac{4n!}{2n!}=2^{2n}[1\cdot 3\cdot \mathrm{5...}(4n-1\left)\right]$

$\frac{{}^{4n}{C}_{2n}}{{}^{2n}{C}_n}=\frac{\frac{4n!}{2n!2n!}}{\frac{2n!}{n!n!}}=\frac{4n!/2n!}{(2n!/n!{)}^2}=\frac{1\cdot 3\cdot 5\cdots (4n-1)}{[1\cdot 3\cdot 5\cdots (2n-1){]}^2}$