Prove that ${^{2n}}{C}_{n} = \dfrac{2^n \times 1 \times 3 \times \cdots \left(2n - 1\right)}{n!}$
$\begin{aligned}
{^{2n}}{C}_{n} & = \dfrac{2n \times \left(2n - 1\right) \times \left(2n - 2\right) \times \left(2n - 3\right) \times \left(2n - 4\right) \times \cdots \times 3 \times 2 \times 1}{\left(2n - n\right)! \times n!} \\\\
& = \dfrac{\left[2n \times \left(2n - 2\right) \times \left(2n - 4\right) \times \cdots \times 2\right] \left[\left(2n - 1\right) \times \left(2n - 3\right) \times \cdots \times 3 \times 1\right]}{n! \times n!} \\\\
& = \dfrac{2^{n} \left[n \times \left(n - 1\right) \times \left(n - 2\right) \times \cdots \times 1\right] \left[\left(2n - 1\right) \times \left(2n - 3\right) \times \cdots \times 3 \times 1\right]}{n! \times n!} \\\\
& = \dfrac{2^{n} \times n! \times \left[\left(2n - 1\right) \times \left(2n - 3\right) \times \cdots \times 3 \times 1\right]}{n! \times n!} \\\\
& = \dfrac{2^{n} \times \left(2n - 1\right) \times \left(2n - 3\right) \times \cdots \times 3 \times 1}{n!}
\end{aligned}$
Hence proved.