Simplifying a binomial coefficient summation I am trying to solve a problem related to expected value, but I am having problem trying to reduce a binomial coefficient expression.
How can I prove the following?
Assume $a \le k \le n$,
$$\sum_{i=0}^{k}\frac{{ i \cdot  {{k}\choose{i}}}\cdot {{n-k}\choose{a-i}}} {{{n}\choose{a}}} 
 = \frac{a\cdot k}{n}$$

\begin{align}
\sum_{i=0}^{k}\frac{{ i \binom{k}{i}} {{n-k}\choose{a-i}}} {\binom{n}{a}} 
&= \sum_{i=1}^{k}\frac{{ i \binom{k}{i}} {{n-k}\choose{a-i}}} {\binom{n}{a}} \\
&= \frac{\sum_{i=1}^{k}{ i \binom{k}{i}} {{n-k}\choose{a-i}}} {\binom{n}{a}} \\ 
&= \frac{\sum_{i=1}^{k}{ i  \frac{k}{i} \binom{k-1}{i-1}} {{n-k}\choose{a-i}}} {\frac{n}{a}\binom{n-1}{a-1}} \\ 
&= \frac{ak}{n}\cdot \frac{\sum_{i=1}^{k}{\binom{k-1}{i-1}} {{n-k}\choose{a-i}}} {\binom{n-1}{a-1}} \\ 
&= \frac{ak}{n} &&\text{by Vandermonde's identity}
\end{align}
