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Just an algebraic step within the well known solution for the number of triangulations of a convex polygon!

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$$\begin{align*} \frac{1}{n-1}\binom{2n-4}{n-2}&=\frac{(2n-4)!}{(n-1)!(n-2)!}\\\\ &=\frac{\Big((2n-4)(2n-6)\dots(2)\Big)\Big((2n-5)(2n-7)\dots(3)(1)\Big)}{(n-1)!(n-2)!}\\\\ &=\frac{2^{n-2}(n-2)!\Big((2n-5)(2n-7)\dots(3)(1)\Big)}{(n-1)!(n-2)!}\\\\ &=2^{n-2}\frac{(2n-5)(2n-7)\dots(3)(1)}{(n-1)!} \end{align*}$$

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Thank you Brian! – Félix Jan 25 '13 at 20:25
@Felix: You’re welcome! – Brian M. Scott Jan 25 '13 at 23:16

First, rewrite the denominator as $(n-1)\cdot(n-2)!$, then multiply numerator and denominator by $(n-2)!$ :


Next, distribute the $n-2$ factors of $2$ over the $n-2$ 'elements' of $(n-2)!$ in the numerator:


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.

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Thanks a lot Steven! – Félix Jan 25 '13 at 20:11

\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}

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