Standard Normal Moments and Combinatorics At around 16-17 mins in this video, the professor calculates the even moments of the standard normal. If $Z \thicksim N(0,1)$ then $$\mathbb{E}[Z^{2n}] = \frac{(2n)!}{2^n \cdot n!}.$$ The right hand side is the number of ways to form $n$ partnerships out of $2n$ people and hints that the moments are linked to this combinatorics problem.
My question is, how are these linked? Not entirely sure how to approach this. A possible approach I've seen in another post is expressing $\mathbb{E}[Z^{2n}]$ recursively. Although, the connection is not apparent (at least for me).
 A: They are linked via the moment generating function. Namely we have for any $Z$
$$E[Z^n]=\frac{d^n}{dt^n} E[e^{tZ}] (0).$$
For standard Gaussian:
$$E[e^{tZ}]=\int C e^{tx}e^{-x^2/2}dx=\int C e^{(t^2-(x-t)^2)/2}dx=e^{-t^2/2}$$
Now
$$E[Z^n]=\frac{d^n}{dt^n}e^{t^2/2}(0).$$
All of this is also explained in the video.
We claim that this is equal to the number of ways to pair up $n$ objects (in particular, is zero if $n$ is odd).  Spoiler: the "objects" are the differentiations.
The key idea is to think about differentiating $n$ times turn by turn. The first of the $n$
differentiations must act ”upstairs” - that is by differentiating $t^2/2$ in the exponent. This
”brings down” a factor of $t$ (since $\frac{d}{dt} e^{t^2}/2
= te^{t^2/2}$) . The second differentiation, by the product rule, can act either
”upstairs” by differentiating $t^2/2$ in the exponent again, bringing down another factor
of $t$ (for a term of $t^2e^{t^2/2}$), or ”downstairs”, differentiating the $t$-factor that the first differentiation
created (for a term of $e^{t^2/2}$). Then, for each of those,  the third differentiation can either create
yet another factor of $t$ (for terms $t^3e^{t^2/2}$ and $te^{t^2/2}$), or erase either one of the two previously created factors
(from $t^2e^{t^2/2}$ we get $2te^{t^2/2}$ - one for the "first" $t$ and one for the "second" $t$; there is no term for $e^{t^2/2}$  since no "downstairs" $t$ is available to erase). We can think of each ”upstairs” differentiation as ”creating” a factor of $t$ (we
can give it an extra name of ”$t$ created at the i-th differentiation” for bookkeeping purposes,
though it is of course the same variable $t$ - the only variable in this computation);
and each ”downstairs” differentiation as picking one of the factors of $t$ created earlier, and
erasing it. All this choosing ”up” or ”down” and creating and erasing factor goes on for $n$
turns (if no ”down” factors are available at some point, then the ”upstairs” differentiation
must be chosen), at which point the complete $n$-th derivative is computed.
We note that in this way some of the derivative steps are paired off – if the $j$th differentiation
acted downstairs to erase the factor of $t$ created by the $i$th differentiation, then
we think of $i$th and $j$th differentiation as paired. In this way of thinking, the unpaired
differentiations must all act upstairs and create factors of t that remain in the final result.
Of course, if we then evaluate the resulting derivative at $t = 0$, any term containing
unpaired differentiation (and hence a positive power of $t$) will vanish! The terms that
will remain will be precisely those which came from the ”fully paired” differentiation
sequences, with each of those contributing the factor of $e^{t^2/2}|_{t=0} = 1$. Thus the value of
our $n$th derivative at $t = 0$ is simply the number of ways of pairing up the $n$ differentiations.
Remarks: This works for higher dimensional Gaussians as well, giving, among other things the Isserlis's theorem. In fact, this is a  of 0-D version of Feynman's path integrals (here the space-time is a point,  and we are integrating over the space of functions i.e. over $\mathbb{R}$, the action $S$ is quadratic $x^2/2$ (this is a "free" or "non-interacting" theory), and the path integral of some function of the field $f(x)$ is $\int_\mathbb{R} f(x) exp(-S(x))dx$; we get that the answer is given by summing over univalent graphs on vertices coming from $f$. For more interesting theories, with action having, say, cubic terms, we would have to sum over trivalent graphs - which now can have internal vertices. Pursuing this in full generality should lead to Feynman diagrams.
