Evaluate the summation involving binomials. $\sum _{ i=0 }^{ 100 }{\binom{k}{i}}*{\binom{M-k}{100-i}*\frac{k-i}{M-100}}/{\binom{M}{100}}$
I wrote the first few terms but couldn't find any pattern and how to club the terms. Help.
 A: Let's generalize a bit by introducing indices:
$$
\sum_i \binom{k}{i}
         \cdot \binom{M - k}{r - i}
         \cdot \frac{k - i}{M - r}
         / \binom{M}{r}
 = \frac{1}{(M - r) \binom{M}{r}} 
     \cdot \sum_i \binom{k}{i} \binom{M - k}{r - i} \cdot (k - i)
$$
We can split the $k - i$ to simplify.
So we are interested in sums of the forms:
$$
\sum_i \binom{a}{i} \binom{b}{r - i}
$$
and
$$
\sum_i i \binom{a}{i} \binom{b}{r - i}
$$
Now remember:
$$
(1 + z)^a = \sum_i \binom{a}{i} z^i \\
z \frac{\mathrm{d}}{\mathrm{d} z} (1 + z)^a = \sum_i i \binom{a}{i} z^i
$$
and also:
$$
\left(\sum_i u_i \right) \cdot \left( \sum_i v_i \right)
  = \sum_i \sum_{0 \le j \le i} u_j v_{i - j}
$$
In particular:
\begin{align}
\sum_{r \ge 0} \sum_i \binom{a}{i} \binom{b}{r - i} z^r
 &= \left(\sum_i \binom{a}{i} z^i \right)
      \cdot \left(\sum_i \binom{b}{i} z^i \right) \\
 &= (1 + z)^a \cdot (1 + z)^b \\
 &= (1 + z)^{a + b}
\end{align}
Comparing coefficients you get Vandermonde's convolution:
$$
\sum_i \binom{a}{i} \binom{b}{r - i} = \binom{a + b}{r}
$$
For the other half:
\begin{align}
\sum_r \sum_i i \binom{a}{i} \binom{b}{r - i} z^r
   &= \left(\sum_i i \binom{a}{i} z^i \right)
        \cdot \left(\sum_i i \binom{b}{i} z^i \right) \\
   &= a z (1 + z)^{a - 1} \cdot (1 + z)^b \\
   &= a z (1 + z)^{a + b - 1}
\end{align}
The coefficient of $z^r$ here is a bit trickier:
$$
[z^r] a z (1 + z)^{a + b - 1}
  = a [z^{r - 1}] (1 + z)^{a + b - 1}
  = a \binom{a + b - 1}{r - 1}
$$
I'm sure you can take it from here.
A: It's easy
\begin{align*}
\frac{1}{\binom{M}{100}}\sum_{i=0}^{100}\frac{k-i}{M-100}\binom{k}{i}\binom{M-k}{100-i} &= \frac{k}{M\binom{M-1}{100}}\sum_{i=0}^{100}\binom{k-1}{i}\binom{M-k}{100-i}\\
&= \frac{k}{M\binom{M-1}{100}}\binom{M-1}{100}\\
&= \frac{k}{M}
\end{align*}
