How to transform $2^{n-2}\frac{(2n-5)(2n-7)...(3)(1)}{(n-1)(n-2)...(3)(2)(1)}$ into $\frac{1}{n-1}\binom{2n-4}{n-2}$? Just an algebraic step within the well known solution for the number of triangulations of a convex polygon!
 A: $$\begin{align*}
\frac{1}{n-1}\binom{2n-4}{n-2}&=\frac{(2n-4)!}{(n-1)!(n-2)!}\\\\
&=\frac{\Big((2n-4)(2n-6)\dots(2)\Big)\Big((2n-5)(2n-7)\dots(3)(1)\Big)}{(n-1)!(n-2)!}\\\\
&=\frac{2^{n-2}(n-2)!\Big((2n-5)(2n-7)\dots(3)(1)\Big)}{(n-1)!(n-2)!}\\\\
&=2^{n-2}\frac{(2n-5)(2n-7)\dots(3)(1)}{(n-1)!}
\end{align*}$$
A: First, rewrite the denominator as $(n-1)\cdot(n-2)!$, then multiply numerator and denominator by $(n-2)!$ :
$$2^{n-2}\frac{(2n-5)(2n-7)\ldots(3)(1)(n-2)(n-3)\ldots(2)(1)}{(n-1)(n-2)!(n-2)!}$$
Next, distribute the $n-2$ factors of $2$ over the $n-2$ 'elements' of $(n-2)!$ in the numerator:
$$\frac{(2n-5)(2n-7)\ldots(3)(1)(2n-4)(2n-6)\ldots(4)(2)}{(n-1)(n-2)!(n-2)!}$$
Now, just interleave the even and odd values in the numerator:
$$\frac{(2n-4)(2n-5)(2n-6)(2n-7)\ldots(3)(2)(1)}{(n-1)(n-2)!(n-2)!}$$
But the numerator is just $(2n-4)!$, so the overall expression is $\dfrac{(2n-4)!}{(n-1)(n-2)!(n-2)!}$, which is exactly your second expression.
A: \begin{eqnarray}
2^{n-2}\frac{(2n-5)(2n-7)\ldots3\cdot1}{(n-1)(n-2)\ldots3\cdot2\cdot1}&=&
2^{n-2}\frac{(2n-4)(2n-5)\ldots\cdot3\cdot2\cdot1}{[(n-1)(n-2)\ldots2\cdot1][(2n-4)(2n-6)\ldots4\cdot2]}\\
&=&2^{n-2}\frac{(2n-4)!}{[(n-1)(n-2)\ldots2\cdot1][2^{n-2}(n-2)(n-3)\ldots2\cdot1]}\\
&=&\frac{(2n-4)!}{(n-1)!(n-2)!}=\frac{1}{n-1}\cdot\frac{(2n-4)!}{[(n-2)!]^2}\\
&=&\frac{1}{n-1}{2n-4\choose n-2}.
\end{eqnarray}
