TITLE: How to calculate the $4$th central moment of binomial distribution?
QUESTION [8 upvotes]: I just derived it by using the generation function to first get raw moments. The result is 

$(-1+3np^2-6p^2-3np+6p)n(p-1)p$.

It was merely brutal force calculation, nothing interesting.  So I was wondering, if there any one knows tricks that could simplify the process a bit.

REPLY [2 votes]: If you decide to pursue Dilips' strategy: For $k\geq 0$, define
$$m_k=E\!\left[{X\choose k}\right]=\sum_{x\geq 0} {x\choose k} {n\choose x} p^x (1-p)^{n-x}.$$
We may not know what these numbers are, but we do know that
\begin{eqnarray*}
\sum_{k\geq0} m_k y^k &=&  \sum_{k\geq0} \sum_{x\geq 0} {x\choose k} {n\choose x} p^x (1-p)^{n-x} y^k \\[5pt]
& =&  \sum_{x\geq 0} (1+y)^x {n\choose x} p^x (1-p)^{n-x}\\[5pt] &=& (py+1)^n.
\end{eqnarray*}
Extracting the coefficient of $y^k$ on both sides gives
$$m_k=E\!\left[{X\choose k}\right]= {n\choose k} p^k.$$