Factorial simplification help; how did wolfram get this?

Question

in this problem n is assumed to be odd: Here's what I am trying to simplify: $$ \frac {(n-2)!( \frac {n-1}{2})!} { (n-1)! (\frac {n-3} {2}) !} $$

Wolfram is telling me that this should simplify to $$\frac {1}{2} $$ but I am very confused about how it gets there. I understand that the two factorial terms combine, but then I have $$ (n-2) (\frac {n-1} {2}) $$ as my final simplification and I really have no idea how to move from that to $$\frac{1}{2}$$ Any help would be appreciated.

Answer 1

First, note that

$$\frac{(n-2)!}{(n-1)!} = \frac{(n-2)(n-3)(n-4) \cdots (2)(1)}{(n-1)(n-2)(n-3)(n-4) \cdots (2)(1)} = \frac{1}{n-1}$$

Similarly, notice that

$$\dfrac{ \left( \dfrac{n-1}{2} \right)! }{\left( \dfrac{n-3}{2} \right)!} = \dfrac{ \left( \dfrac{n-1}{2} \right)\left( \dfrac{n-3}{2} \right)\left( \dfrac{n-5}{2} \right)\cdots(2)(1) }{\left( \dfrac{n-3}{2} \right)\left( \dfrac{n-5}{2} \right)\cdots(2)(1) } = \frac{n-1}{2}$$

Multiply the two together for the desired result.

Answer 2

Here is a variation based upon the functional equation: $n!=n(n-1)!$.

We obtain \begin{align*} \color{blue}{\frac {(n-2)!( \frac {n-1}{2})!} { (n-1)! (\frac {n-3} {2}) !}} &=\frac {(n-2)!}{(n-1)(n-2)!}\cdot \frac {\frac{n-1}{2}\left(\frac{n-3}{2}\right)!}{\left(\frac{n-3}{2}\right)!}\\ &=\frac{1}{n-1}\cdot\frac{n-1}{2}\\ &\,\,\color{blue}{=\frac{1}{2}} \end{align*} and the claim follows.