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.

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 $p$.
