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.

Recall that if $X\sim\mathrm{Bin}(n,p)$, then $\mathbb E[X] = np$ and $\mathrm{Var}(X)=np(1-p)$. Given $\mathbb E[X] = 4$ and $\mathrm{Var}(X) = 3$, we have $np = 4$ and $np(1-p)=3$. Hence $n=16$, $p=\frac14$. So the distribution of $X$ is given by
$$\mathbb P(X=k) = \binom {16}k \left(\frac14\right)^k\left(\frac34\right)^{16-k}, k=0,1,\ldots,16.$$
The second moment of $X$ is
$$\mathbb E[X^2] = \mathrm{Var}(X) + \mathbb E[X]^2 = 3 + 4^2 = 19.$$
The generating function of $X$ is
$$\begin{align*}
\mathbb E[s^X] &= \sum_{k=0}^\infty\mathbb P(X=k)s^k.
\end{align*}$$
Since $X$ is the sum of $16$ i.i.d. $\mathrm{Ber}(p)$ random variables, each with generating function $1-p+ps$, we have
$$\mathbb E[s^X] := P(s) = (1-p+ps)^n = \left(\frac34 + \frac14s\right)^{16}.$$
The third central moment of $X$ is
$$\begin{align*}
\mathbb E[(X-\mathbb E[X])^3] &= \mathbb E[(X-4)^3]\\ 
&= \mathbb E[X^3 - 12X^2 + 48X - 64]\\
&= \mathbb E[X(X-1)(X-2) - 9X^2 + 46X - 64]\\
&= \mathbb E[X(X-1)(X-2)] - 9\mathbb E[X^2] + 46\mathbb E[X] - 64.
\end{align*}$$
Now,
$$\begin{align*}\mathbb E[X(X-1)(X-2)] &= \lim_{s\uparrow1}P^{(3)}(s)\\ 
&= \lim_{s\uparrow1}\left(\frac14\right)^3 16\cdot15\cdot14\left(\frac34+\frac14s\right)^{13}\\
&= \frac{105}2. \end{align*}$$
Hence
$$\begin{align*}
\mathbb E[(X-\mathbb E[X])^3] &= \mathbb E[X(X-1)(X-2)] - 9\mathbb E[X^2] + 46\mathbb E[X] - 64\\
&= \frac{105}2 - 9\cdot19 + 46\cdot 4 - 64\\
&= \frac32.
\end{align*}$$
