How to calculate the $4$th central moment of binomial distribution? 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.
 A: Let $X$ be binomial with $n$ trials and success probability $p$. I assume by fourth central moment you mean $E[(X - np)^4]$. Write $X = \sum_{i = 1} ^ n X_i$ where the $X_i$ are iid Bernoulli with success probability $p$. Then we want $$E\left[\left(\sum_{i = 1} ^ n (X_i - p)\right)^4 \right] = E\left[\sum_{1 \le i, j, k, l \le n} (X_i - p)(X_j - p)(X_k - p)(X_l - p) \right] = \sum_{1 \le i, j, k, l \le n}E[(X_i - p)(X_j - p)(X_k - p)(X_l - p)]$$ which effectively turns this into a counting problem. The terms in the sum on the right hand side are $0$ unless either $i = j = k = l$, which happens $n$ times, or when there are two pairs of matching indicies, which happens $\binom n 2 \binom 4 2 = 3n(n - 1)$ times. This gives us $$E[(X - np)^4] = n\mu_4 + 3n(n - 1)\sigma^4 \tag{$\dagger$}$$ as the fourth central moment, when $\sigma^2 = p(1 - p)$ is the variance $X_1$ and $\mu_4 = p(1 - p)^4 + p^4 (1 - p)$ is the fourth central moment of $X_1$. Note that in deriving ($\dagger$) we actually derived a general formula for the fourth central moment of a sum of iid random variables.
A: An alternative approach would be to use cumulants. 
The fourth central moment of a random variable $X$ can be 
expressed in terms of cumulants as follows:
$$\mu_4(X)=\kappa_4(X)+3\kappa^2_2(X).$$
Now, cumulants add over independent random variables and 
the second cumulant is just the variance, i.e., $\kappa_2=\mu_2$.
Writing $Y=\sum_{i=1}^n Z_i$, where 
the $Z_i\,$s are i.i.d. random variables, we have 
\begin{eqnarray*}
\mu_4(Y)&=&\kappa_4(Y)+3\kappa^2_2(Y)\\
&=&n\kappa_4(Z)+3[n\kappa_2(Z)]^2\\
&=&n\left[\mu_4(Z)-3\kappa_2^2(Z)\right]+3[n\kappa_2(Z)]^2\\
&=&n\, \mu_4(Z) +3n(n-1)\,\mu_2^2(Z).
\end{eqnarray*}  
A: 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
$$m_k=E\!\left[{X\choose k}\right]= {n\choose k} p^k.$$
