Binomial distribution central moment calculation 
If for a binomial distribution the mean is $4$ and variance is $3$, find th $3^{\text{rd}}$ central moment. 

I understand that the first and second central moments are mean and variance respectively, but what is the $3^{\text{rd}}$? A Google search only reveals formulas with $n$ in them, which is information that is, as per the question, unavailable. The answer to the  question is $3/2$, but I can't figure out where it comes from.

By definition, the third central moment of a random variable $X$ is $E[(X-\mu)^3]$ where $\mu = E[X]$.  
The Binomial($n,p$) random variable $X$ can be written as the sum of $n$ independent Bernoulli($p$) random variables $Y_i$, but we want to subtract the mean, so write 
$ X - np = \sum_{j=1}^n Z_j$ where $$Z_j = Y_j - p = \cases{1-p & with probability $p$\cr -p & with probability $1-p$\cr}$$
are independent.  Now
$$ E[(X - np)^3] = \sum_{i=1}^n \sum_{j=1}^n \sum_{k=1}^n E[Z_i Z_j Z_k]$$
There are $n$ terms where $i=j=k$, and $E[Z_i^3] = p (1-p)^3 - p^3 (1-p)$.
The other terms are either of the form $E[Z_i Z_j^2]$ with $i \ne j$ or $E[Z_i Z_j Z_k]$ with $i,j,k$ all different, and those are all $0$ since 
the $Z$'s are independent and have mean $0$.  So we conclude that the 
third central moment is 
$$ E[(X -np)^3] = n (p (1-p)^3 - p^3 (1-p))$$
which can be simplified to $n p (1-p)(1-2p)$.
