$r^{th}$ factorial of Hypergeometric Distribution $r^{th}$ factorial
$E(X_r)=E[X(X-1)\dots(X-r+1)]$ in case of Hyper-geometric Distribution.
$$f_X(x) = \left\{ \begin{array}{ll}
         \frac{\binom{a}{x}\binom{N-a}{N-x}}{\binom{N}{n}} &  x \in S_x\{\max(0,n-N+a,\dots,\max(n,a)\}\\
        0 & otherwise \end{array} \right.  $$
I am getting problem in solving $E(X_r)$

Particular step:
How
$$\sum_{k=\max(0,n-N+a-r)}^{\min(n-r,a-r)}
 \binom{a-r}{k}\binom{(N-r)-(a-r)}{(n-r)-k}=\binom{N-r}{n-r}$$

 A: Let us change notation. Let $a-r=b$ and $(N-r)-(a-r)=g$. Let $n-r=p$. Then the identity you are trying to prove can be rewritten as
$$\sum_{k=0}^p \binom{b}{k}\binom{g}{p-k}\overset{?}{=} \binom{b+g}{p}.\tag{$1$}$$
There are $b$ boys and $g$ girls in a class. In how many ways can we form a committee of $p$ people? Obviously $\binom{b+g}{p}$.
Let us count the number of committees another way. We could choose $k=0$ boys and $p-0$ girls; or we could choose $k=1$ boys and $p-1$ girls. Or we could choose $2$ boys and $p-2$ girls, and so on. The sum on the left of $(1)$represents that way of counting the committees. The two ways of counting must give the same answer, and hence the desired identity holds.
Remark: One might worry about what happens if, for example, there is a total of $5$ boys in the class, and $30$ girls, and we want a committee of say $8$ people. Then we have two choices. We can restrict $k$ to run from $0$ to $5$. That is essentially the notational approach taken in the material quoted. Or else we can define $\binom{x}{y}$ to be $0$ if $x$ and $y$ are non-negative integers such that $x\lt y$. With either approach, the identity of $(1)$ is correct for all possible $b$, $g$, and $p$. 
