Question

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

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

Answer 1

Answer: (C)

$1\cdot 3\cdot 5....\cdot (2n-1)$

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