Question

The product of first $n$ odd natural numbers equals to

• A. ${}^{2n}C_n\times {}^n{P}_n$
• B. ${\left(\frac{1}{2}\right)}^n{}^{2n}C_n\times {}^n{P}_n$
• C. ${\left(\frac{1}{4}\right)}^n{}^{2n}C_n\times {}^{2n}{P}_n$
• D. none of these

Answer 1

The correct option is B

${\left(\frac{1}{2}\right)}^n{}^{2n}C_n\times {}^n{P}_n$

Explanation for correct option:

Product of first $n$ odd natural numbers$=1·3·5............(2n-1)$

Multiply and divide by $2·4·6·......\left(2n\right)$

$$\begin{gathered} =1·3·5·\dots ·(2n-1)\times \frac{2·4·6·\dots ·\left(2n\right)}{2·4·6·\dots ·\left(2n\right)} \\ =\frac{1·2·3·4·5·\dots ·(2n-1)·\left(2n\right)}{2^n\left[1·2·\dots ·\left(n\right)\right]} \\ =\frac{\left(2n\right)!}{2^{n}n!} \\ =\frac{1}{2^n}\frac{\left(2n\right)!}{n!}\times \frac{n!}{n!} \\ =\frac{1}{2^n}\frac{\left(2n\right)!}{n!\times n!}\times \frac{n!}{1} \\ =\frac{1}{2^n}\frac{\left(2n\right)!}{n!\times n!}\times \frac{n!}{0!} \\ =\frac{1}{2^n}\frac{\left(2n\right)!}{n!\times \left(2n-n\right)!}\times \frac{n!}{(n-n)!} \\ ={\left(\frac{1}{2}\right)}^n{}^{2n}{C}_n{}^n{P}_n \end{gathered}$$

Hence, option (B) is correct.