Question

If $n$ is any positive integer, then $\frac{1}{2^n}\left({}^{2n}{P}_n\right)$ is equal to

• A. $2\cdot 4\cdot 6\cdots \cdot \left(2n\right)$
• B. $1\cdot 2\cdot 3\cdots \cdot n$
• C. $1\cdot 3\cdot 5\cdots \cdot \left(2n-1\right)$
• D. $1\cdot 2\cdot 3\cdots \cdot \left(3n\right)$
• E. $2\cdot 4\cdot 6\cdots \cdot \left(2n+2\right)$

Answer 1

Answer: (C)

$\therefore \frac{1}{2^n}\left({}^{2n}{P}_n\right)=\frac{1}{2^n}\left(\frac{2n!}{n!}\right)$
$=\frac{2n(2n-1)(2n-2)(2n-3)\dots 3\cdot 2\cdot 1}{2^{n}n!}$
$2^n\times n(n-1)(n-2)\dots 3\times 2\times 1$
$=\frac{[1\cdot 3\cdot 5\dots (2n-1)]}{2^{n}n!}$
$=1\cdot 3\cdot 5\dots (2n-1)$