Question

Which of the following is equal to $\frac{\mathrm{1.3.5...}(2n-1)}{\mathrm{2.4.6...}\left(2n\right)}$?

• A. $\left(2n\right)!÷{\left(2^n(n!)\right)}^2$
• B. $\left(2n\right)!÷n!$
• C. $(2n-1)÷(n-1)$
• D. $2^n$

Answer 1

The correct option is A $\left(2n\right)!÷{\left(2^n(n!)\right)}^2$

$\frac{\mathrm{1.3.5..}(2n-1)}{\mathrm{2.4.6..}\left(2n\right)}=\frac{\mathrm{1.2.3...}(2n-1)\left(2n\right)}{\left[\mathrm{2.4.6..}\right(2n){]}^2}$

$\Rightarrow \frac{2n!}{[2^{n}n!{]}^2}=\frac{2n!}{2^{2n}(n!{)}^2}$