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 $a \le k \le n$,
$$\sum_{i=0}^{k}\frac{{ i \cdot  {{k}\choose{i}}}\cdot {{n-k}\choose{a-i}}} {{{n}\choose{a}}} 
 = \frac{a\cdot k}{n}$$

We can do double counting.
Suppose you wish to select a committee of $a$ people out of $k$ adults and $n-k$ kids. You also need to choose one of the adults in the committee to be the leader. We can do this by selecting a number of adults, say $i$ out of the available $k$ adults, and select one out of these $i$ adult committee member to be the leader. Then we need to fill the committee by selecting $a-i$ out of the available $n-k$ kids. The number of way to do this is given by the following:
$$
\sum_{i=0}^{k}i\cdot\binom{k}{i}\cdot\binom{n-k}{a-i}
$$
Another way to form the committee is to first select one leader out of $k$ adults, and then select $a-1$ members out of combined $n-1$ remaining adults and kids. The number of way to do this is given below.
$$
k\cdot\binom{n-1}{a-1}
$$
Because both expressions are used to count the same thing, they must be equal.
$$
\sum_{i=0}^{k}i\cdot\binom{k}{i}\cdot\binom{n-k}{a-i}
=
k\cdot\binom{n-1}{a-1}
$$
Now simply divide both side by $\binom{n}{a}$ to get your initial expression
