Given that $X \sim \operatorname{Binomial}(n,p)$, Find $\mathbb{E}[X(X-1)(X-2)(X-3)]$ 
Given that $X \sim \operatorname{Binomial}(n,p)$, Find $\mathbb{E}[X(X-1)(X-2)(X-3)]$.

It is suggested that I can transform it into 
\begin{align}
\mathbb{E}[X(X-1)(X-2)(X-3)] 
&=\sum_{k=0}^n k(k-1)(k-2)\mathbb{P}\{X=k\}\\
&=\sum_{k=3}^{n+3} (k-3)(k-4)(k-5)\mathbb{P}\{X=k-3\}\\
&=\sum_{k=0}^n i(i-1)(i-2)\mathbb{P}\{X=i\}
\end{align}
But then I just have no idea about how can i do it. I suspect that it needs something similar to this post but the steps are quite different from this one.
Please help.
 A: One way is to use characteristic functions. The characteristic function of $B(n,p)$ is  $\phi (t)=(pe^{it}+(1-p))^{n}$. The moments of $X$ are given by $i^{n}EX^{n}=\phi^{(n)}(0)$. You can compute the first 4 moments of $X$ using this and then use the expanded form  of $X(X_1)(X-2)(X-3)$.
A: Hint: Probability Generating Function (see https://en.wikipedia.org/wiki/Probability-generating_function)
Look at the properties section, the part where it talks about the kth factorial moment.
A: Look at  $f(s)= E [ s^X]$. 
1) Observe that
$$f(s) = (sp + (1-p))^n=(1+ (s-1)p)^n =\sum_{k=0}^n \binom{n}{k}p^k (s-1)^k=\sum_{k=0}^n \frac{n!}{(n-k)!} \frac{(s-1)^k}{k!},$$ 
the RHS identified as the Taylor series of $f$ about $s=1$.  
2) Differentiating under the expectation (which is just a finite sum here),  we obtain $$f'(s) = E[ X s^{X-1}] ,~f''(s) = E[X (X-1) s^{X-2}],\dots.$$
You're looking for $f^{(4)}(1)$. 
A: Start as suggested, and write down what the probability mass function (pmf) of the Binomial actually is:
$$\begin{align*}
\mathbb{E}[X(X-1)(X-2)(X-3)]
&= \sum_{k=0}^n k(k-1)(k-2)(k-3)\mathbb{P}\{X=k\}\\
&= \sum_{k=4}^n k(k-1)(k-2)(k-3)\mathbb{P}\{X=k\}\\
&= \sum_{k=4}^n k(k-1)(k-2)(k-3)\binom{n}{k}p^k(1-p)^{n-k}\\
&= \sum_{k=4}^n k(k-1)(k-2)(k-3)\frac{n!}{k!(n-k)!}p^k(1-p)^{n-k}\\
&= \sum_{k=4}^n \frac{n!}{(k-4)!(n-k)!}p^k(1-p)^{n-k}\\
&= \sum_{k=4}^n \frac{n(n-1)(n-2)(n-3)(n-4)!}{(k-4)!((n-4)-(k-4))!}p^k(1-p)^{n-k}\\
&= n(n-1)(n-2)(n-3)p^4\sum_{k=4}^n \binom{n-4}{k-4}p^{k-4}(1-p)^{(n-4)-(k-4)}\\
&= n(n-1)(n-2)(n-3)p^4\sum_{\ell=0}^n \binom{n-4}{\ell}p^{\ell}(1-p)^{(n-4)-\ell}\\
&= \boxed{n(n-1)(n-2)(n-3)p^4}
\end{align*}$$
since $\sum_{\ell=0}^n \binom{n-4}{\ell}p^{\ell}(1-p)^{(n-4)-\ell}=1$, recognizing the sum of probabilities for a Binomial with parameters $n-4$ and $p$.
