Question

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

• A. $2\cdot4\cdot6 \cdots \cdot\left(2n\right)$
• B. $1\cdot2\cdot3 \cdots \cdot n$
• C. $1\cdot 3\cdot 5 \cdots \cdot\left(2n-1\right)$
• D. $1\cdot2\cdot3 \cdots \cdot\left(3n\right)$
• E. $2\cdot4\cdot6 \cdots \cdot\left(2n+2\right)$

Answer 1

Answer: (C)

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