Proving a combinatorial identity without double counting Consider the following problem:

We have a collection of m white and n black balls. We pick k balls (k ≤ m+n) without replacement. What is the expected number of white balls in the sample?

Counting method 1: The ratio of white balls in the collection is m/(m+n). If we repeatedly sample the collection, we expect that on average, our sample has the same ratio. Therefore, the expected number of white balls in a sample of size k is km/(m+n).
As a side note, I think the above argument somehow depends on the unbiasedness of 
the mean estimator. Am I right?

Counting method 2: We directly compute the expected number of white balls in the sample:
$$ \frac{\sum_{i=0}^{k} i\binom{m}{i}\binom{n}{k-i}}{\binom{m+n}{k}} \enspace.$$

So, we reach at the identity:
$$ \sum_{i=0}^{k} i\binom{m}{i}\binom{n}{k-i} = \frac{km}{m+n}\binom{m+n}{k} \enspace.$$

How to prove the above result without double counting?
It seems to be provable using Vandermonde's identity as well as $i\binom{m}{i} = m\binom{m-1}{i-1}$. But is there a simpler approach (which does not depend on extra identities such as Vandermonde's)?

 A: If we are looking to evaluate
$$\sum_{q=0}^k q{m\choose q}{n\choose k-q}
= \sum_{q=1}^k q{m\choose q}{n\choose k-q}
= m \sum_{q=1}^k {m-1\choose q-1}{n\choose k-q}$$
there is a very simple proof using the integral representation
$${n\choose k-q}
= \frac{1}{2\pi i}
\int_{|z|=\epsilon} 
\frac{1}{z^{k-q+1}} (1+z)^n
\; dz.$$
This is zero when $q\gt k$ so we may extend the sum to infinity to get
$$\frac{m}{2\pi i}
\int_{|z|=\epsilon} 
\frac{1}{z^{k+1}} (1+z)^n
\sum_{q\ge 1} {m-1\choose q-1} z^q
\; dz
\\ = \frac{m}{2\pi i}
\int_{|z|=\epsilon} 
\frac{1}{z^{k+1}} (1+z)^n
z (1+z)^{m-1}
\; dz
\\ = \frac{m}{2\pi i}
\int_{|z|=\epsilon} 
\frac{1}{z^{k}} (1+z)^{m+n-1}
\; dz.$$
This is
$$m{m+n-1\choose k-1}
= \frac{km}{m+n} {m+n\choose k}.$$
A: For $i=1$ to $k$, define the indicator random variable $X_i$ by $X_i=1$ if the $i$-th ball picked is white, and by $X_i=0$ otherwise. Then the number $W$ of white balls picked is given by
$$W=X_1+X_2+\cdots+X_k.$$
By the linearity of expectation, we have 
$$E(W)=E(X_1)+E(X_2)+\cdots+E(X_k).\tag{1}$$
For any $i$, the probability that the $i$-th ball picked is white is $\frac{m}{m+n}$. This is because any ball is just as likely as any other to be the obe picked $i$-th. 
Thus $\Pr(X_i=1)=\frac{m}{n}$, and therefore $E(X_i)=\frac{m}{m+n}$. It follows from (1) that $E(W)=k\cdot \frac{m}{m+n}$.
