Let $X$ be Hypergeometric, Find $E\left(\binom{X}{2}\right)$ 
Let X be Hypergeometric: $X \sim \operatorname{HGeom}(w,b,n)$, so that $X$ is the number of white balls in a sample of size $n$ out of a population of $w+b$ white and black balls.
Find $E\left(\binom{X}{2}\right)$ by thinking, without any complicated calculations.

I thought about expanding the sample size $n$ somehow, but I can't write down something meaningful. Do you have an idea?
 A: Not answering directly OP's question - since the following is certainly not "without any complicated calculations". Just to verify computationally the correct result (that another answer reaches intuitively): 
$$\begin{align*}E\left[\dbinom{X}{2}\right]&=E\left[\frac{X(X-1)(X-2)!}{2(X-2)!}\right]=\frac{1}{2}\left(E[X(X-1)]\right)=\frac{1}{2}\left(E[X^2]-E[X]\right)=\\[0.2cm]&=\frac{1}{2}\left(E[X^2]\pm E[X]^2-E[X]\right)=\frac{1}{2}\left(\operatorname{Var}(X)+E[X]^2-E[X]\right)\end{align*}$$ where $\operatorname{Var}(X)$ and $E[X]$ are well known (see here). Thus, substituting yields $$\begin{align*}E\left[\dbinom{X}{2}\right]&=\frac{1}{2}\left(n{w\over w+b}{(w+b-w)\over w+b}{w+b-n\over w+b-1}+\left(n\frac{w}{w+b}\right)^2-n\frac{w}{w+b}\right)=\\[0.2cm]&=\frac{n}{2}\frac{w}{w+b}\left(\frac{b}{w+b}\frac{w+b-n}{w+b-1}+n\frac{w}{w+b}-1\right)=\\[0.2cm]&=\frac{n}{2}\frac{w}{w+b}\left(\frac{b(w+b-n)+nw(w+b-1)-(w+b)(w+b-1)}{(w+b-1)(w+b)}\right)=\\[0.2cm]&=\frac{n}{2}\frac{w}{w+b}\left(\frac{\not bw+\not b^2-bn+nw^2+nwb-nw-w^2-\not bw+w-bw-\not b^2+b}{(w+b-1)(w+b)}\right)=\\[0.2cm]&=\frac{n}{2}\frac{w}{w+b}\left(\frac{(nw-n-w+1)(w+b)}{(w+b-1)(w+b)}\right)=\\[0.2cm]&=\frac{n}{2}\frac{w}{w+b}\left(\frac{(n-1)(w-1)}{(w+b-1)}\right)\end{align*}$$
It is certainly a challenge to derive this result only by thinking.

Assuming that you know the formula for the moments of the hypergeometric distribution $$\begin{align*}E[X]&=n\frac{w}{w+b} \\[0.3cm] E[X^2]&=E[X]\left(\frac{(n-1)(w-1)}{w+b-1}+1\right)\end{align*}$$ you can arrive to this result faster.
A: After taking a sample of $n$ balls focus on unordered pairs
of balls that are present in the sample. 
Give the balls in the sample randomly (i.e. without looking
at their colors) numbers $1,\dots,n$ and for $i<j$ let $X_{i,j}=1$
if the balls numbered $i$ and $j$ are both white and $X_{i,j}=0$
otherwise. 
Then $\binom{X}{2}=\sum_{i<j}X_{i,j}$ so that $\mathbb{E}\binom{X}{2}=\mathbb{E}\sum_{i<j}X_{i,j}=\sum_{i<j}\mathbb{E}X_{i,j}=\binom{n}{2}\frac{w}{w+b}\frac{w-1}{w+b-1}$
The last equality because there are $\binom{n}{2}$ unordered pairs of balls in the sample and for each pair the probability that the balls are both white is $\frac{w}{w+b}\frac{w-1}{w+b-1}$.
