Prove that: ^ 2 n C_n=2^n[1 3 5 (2 n-1)] n! . - Vidyadip

Question

Prove that: ${ }^{2 n} C_n=\frac{2^n[1 \cdot 3 \cdot 5 \ldots \ldots(2 n-1)]}{n!}$.

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Answer

${ }^{2 n} C_n=\frac{2^n[1 \cdot 35 \ldots .(2 n-1)]}{n!}$
$=\frac{(2 n)!}{n!n!}$
$\begin{array}{l}(2 n)(2 n-1)(2 n-2)(2 n-3) \ldots . . \\
=\frac{4 \cdot 3 \cdot 2 \cdot 1}{n!n!} \\
=\frac{\left[2^n \ldots .4 \cdot 2\right][(2 n-1) \ldots \cdot 3 \cdot 1]}{n!n!} \\
=\frac{2^n[1 \cdot 2 \cdot \ldots . n][1 \cdot 3 \cdot 5 \ldots \ldots \cdot(2 n-1)]}{n!n!} \\
=\frac{2^n \times n![1 \cdot 3 \cdot 5 \ldots . \cdot(2 n-1)]}{n!n!} \\
=\frac{2^n[1 \cdot 3 \cdot 5 \ldots \cdot(2 n-1)]}{n!}\end{array}$

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