Question

Whether the given equation $:$ ${}^{4n}{C}_{2n}{:}^{2n}{C}_n=[1\cdot 3\cdot 5\dots \left(4n-1\right)]:[1\cdot 3\cdot 5\dots \left(2n-1\right){]}^2$ ?

• A. True
• B. False

Answer 1

The correct option is A True

$LHS$

$\frac{{}^{4n}{C}_{2n}}{{}^{2n}{C}_n}\Rightarrow \frac{4n!}{(2n!)(2n!)}\times \frac{n!\times n!}{(2n!)}$

$\Rightarrow \frac{4n!}{(2n!{)}^2}\times \frac{(n!{)}^2}{2n!}=\frac{4n!\times (n!{)}^2}{(2n!{)}^3}$

$\Rightarrow \frac{2^{2n}\times 2n!\left[\mathrm{1.3.5.7.........}\right(4n-1\left)\right]\times (n!{)}^2}{(2n!{)}^3}$

$\Rightarrow \frac{2^{2n}\times 2n![\mathrm{1.3.5}.........(4n-1\left)\right]\times (n!{)}^2}{(2n!).(2n!{)}^2}$

$\Rightarrow \frac{2^{2n}\times (2n!)\left[\mathrm{1.3.5........}\right(4n-1\left)\right]\times (n!{)}^2}{\left(2^n{)}^2\right[\mathrm{1.3.5}........(2n-1){]}^2}$

$\Rightarrow \frac{\mathrm{1.3.5}........(4n-1)}{\left[\mathrm{1.3.5.....}\right(2n-1){]}^2}$

$RHS$ $\therefore LHS=RHS$

$\therefore$ The given equation is True