Combinatorial explanation for $2n$th moment of standard normal distribution? The $2n^\text{th}$ moment of the standard normal distribution $X \sim \text{N}(0, 1)$ is given by:
$$\mathbb{E}(X^{2n}) = \frac{(2n)!}{2^n n!},$$
which is also the number of ways to divide $2n$ people into $n$ distinct pairs.  Is this just a coincidence, or is there some combinatorial explanation for this?
 A: I am not sure if there is a simple way to frame this in combinatorial terms, other than to observe the equivalency that you have already observed.  Nevertheless, a preliminary aspect of framing the result in combinatorial terms is to look at the moment generating function of the standard normal distribution.  The MGF and its Maclaurin expansion are given by:
$$m_Z(t) = \exp \Big( \frac{t^2}{2} \Big) = \sum_{\ell=0}^\infty \frac{t^{2\ell}}{2^\ell \ell!}.$$
To facilitate our analysis, let $\mathscr{C}(k)$ denote the class of sets of $2k$ people divided into $k$ distinct pairs, so we have:
$$|\mathscr{C}(k)| = \frac{(2k)!}{2^k k!}.$$
Differentiating the MGF $2n$ times gives:
$$\begin{aligned}
\frac{d^{2n} m_Z}{dt^{2n}}(t) 
&= \sum_{\ell=0}^\infty \bigg( \frac{d}{dt} \bigg)^{2n} \frac{t^{2\ell}}{2^\ell \ell!} \\[6pt]
&= \sum_{\ell=0}^\infty \frac{1}{2^\ell \ell!} \bigg( \frac{d}{dt} \bigg)^{2n} t^{2\ell} \\[6pt]
&= \sum_{\ell=n}^\infty \frac{(2\ell)_{2n}}{2^\ell \ell!} \cdot t^{2\ell-2n} \\[6pt]
&= \sum_{\ell=0}^\infty \frac{(2n+2\ell)!}{2^{n+\ell} (n+\ell)!} \cdot \frac{t^{\ell}}{(2 \ell)!} \\[6pt]
&= \sum_{\ell=0}^\infty |\mathscr{C}(n+\ell)| \cdot \frac{t^{\ell}}{(2 \ell)!}, \\[6pt]
\end{aligned}$$
and so the $2n$th raw moment is:
$$\mathbb{E}(Z^{2n}) = \frac{d^{2n} m_Z}{dt^{2n}}(0) = |\mathscr{C}(n)|.$$
As you can see, combinatorial terms similar to the one of interest to you occur when we differentiate the moment generating function.  This is a general result pertaining to that particular exponential function, so it is not something restricted to the normal distribution.  
