What's $\frac{(2n)!}{n!}$ equal to?

Question

I chanced upon this problem:

$\textbf{Show that} \hspace{0.2cm}\frac{(2n)!}{n!} = 2^n(1 \times 3 \times 5 \times ... \times (2n - 1)).$

I tried the following, and realised I was wrong! :

$\frac{(2n)!}{n!} = \frac{2n(2n-1)(2n-2)(2n-3)\times ... \times 1}{n(n-1)(n-2)(n-3)\times ... \times 1}$

$= \frac{2(2n-1)(2n-2)(2n-3)\times ... \times 1}{(n-1)(n-2)(n-3)\times ... \times 1}$

$= 2(2n-1)(2n-3)(2n-5)\times ... \times 1$

$= 2(1 \times 3 t\times 5 \times ... \times (2n-1))$

Where am I going wrong?

Thanks in advance.

Answer 1

You can write it so: $$\binom{2n}{n}n!$$ By your way we obtain: $$\frac{(2n)1}{n!}=\frac{2n(2n-1)(2n-2)(2n-3)(2n-4)...1}{n(n-1)(n-2)...1}=$$ $$=2(2n-1)2(2n-3)2(2n-5)...=2^n(2n-1)(2n-3)...1=2^{n}(2n-1)!!$$

Answer 2

Your mistake is that you did not collect the $n$ factors $2$ which would lead to the expression $2^n$ in front.

Writing it as follows may clarify this:

$$\frac{2n!}{n!}=\frac{\prod_{k=1}^{2n}k}{\prod_{k=1}^n k} = \frac{\prod_{i=1}^{n} 2i \cdot \prod_{i=1}^{n}(2i-1) }{\prod_{k=1}^{n}k} = \color{blue}{2^n}\frac{\prod_{i=1}^{n} i \cdot \prod_{i=1}^{n}(2i-1) }{\prod_{k=1}^{n}k}$$ $$= 2^n\prod_{i=1}^{n}(2i-1)$$