Question

prove that ${}^{4n}{C}_{2n}{:}^{2n}{C}_n=\mathrm{1.3.5...}\left(4n-1\right):[\mathrm{1.3.5...}\left(2n-1\right){]}^2.$

Answer 1

$\frac{2h!}{n!}=\frac{2n(2n-1)(2n-2)\dots \dots \dots ...3.2)}{n!}$

$=\frac{\left(\mathrm{2.4.6.....}\right(2n-2\left)\right(2n\left)\right)\left(\mathrm{1.3.5.....}\right(2n+1)}{n!}$

$\frac{2n!}{n!}=\frac{2^n(\mathrm{1.2.3...}\dots ..(n-1\left)n\right)\left(\mathrm{1.3.5.....}\right(2n+1)}{n!}$

$\frac{2n!}{n!}=\frac{2^{n}n!\left(\mathrm{1.3.5.....}\right(2n-1)}{n!}=2^n\left(\mathrm{1.3.5....}\right(2n+1\left)\right)$ ………$\left(1\right)$

Replacing n by $2n$

$\frac{4n!}{2n!}=2^{2n}\left(\mathrm{1.3.5....}\right(4n-1\left)\right)$ ………..$\left(2\right)$

Now the given ratio can be written as

${}^{4n}{C}_n|{}^{2n}{C}_n=\frac{4n!}{2n!2n!}\times \frac{n!n!}{2n!}$

${}^{4n}{C}_{2n}|{}^{2n}{C}_n=\frac{4n!}{2n!}\times {\left(\frac{n!}{2n!}\right)}^2$

Now using $\left(1\right)$ and $\left(2\right)$

$\frac{{}^{4n}{C}_n}{{}^{2n}{C}_n}=\frac{2^{2n}(\mathrm{1.3.5...}\dots (4n-1\left)\right)}{2^{2n}\left(\mathrm{1.3.5.....}\right(2n-1){)}^2}$

$\frac{{}^{4n}{C}_n}{{}^{2n}{C}_n}=\frac{\left(\mathrm{1.3.5.....}\right(4n-1\left)\right)}{\left(\mathrm{1.3.5.....}\right(2n-1){)}^2}$.