Question

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

Answer 1

$\frac{\frac{4n!}{\left(2n\right)!\left(2n\right)!}}{\frac{2n!}{n!n!}}({\because }^n{C}_r=\frac{n!}{r!(n-r)!})=\frac{(4n!)\times (n!{)}^2}{(2n!)(2n!)\times (2n!{)}^2}=\frac{\left[\mathrm{1.2.3.4...}\right(4n-1\left)\right(4n\left)\right]\left(n{!}^2\right)}{\left(2n\right)!\left[\mathrm{1.2.3.4....}\right(2n-2\left)\right(2n-1\left)\right(2n){]}^2}=\frac{\left[\mathrm{1.3.5....}\right(4n-1\left)\right]\times \left[\mathrm{2.4.6...4}n\right]\times (n!{)}^2}{\left(2n\right)!\left[\mathrm{1.3.5...}\right(2n-1){]}^2\times [\mathrm{2.4.6....}(2n-2)\left(2n\right){]}^2}=\frac{\left[\mathrm{1.3.5...}\right(4n-1\left)\right]\times 2^{2n}\times \left[\mathrm{1.2.3...2}n\right]\times n{!}^2}{\left(2n\right)!\times \left[\mathrm{1.3.5...}\right(2n-1{)}^2\times 2^{2n}\times n{!}^2]}=\frac{\left[\mathrm{1.3.5...}\right(4n-1\left)\right]}{\left[\mathrm{1.3.5...}\right(2n-1){]}^2}$
Hence proved.