<s>[INST] 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!
 [/INST] $$\begin{align*}
\frac{1}{n-1}\binom{2n-4}{n-2}&=\frac{(2n-4)!}{(n-1)!(n-2)!}\\\\
&=\frac{\Big((2n-4)(2n-6)\dots(2)\Big)\Big((2n-5)(2n-7)\dots(3)(1)\Big)}{(n-1)!(n-2)!}\\\\
&=\frac{2^{n-2}(n-2)!\Big((2n-5)(2n-7)\dots(3)(1)\Big)}{(n-1)!(n-2)!}\\\\
&=2^{n-2}\frac{(2n-5)(2n-7)\dots(3)(1)}{(n-1)!}
\end{align*}$$
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