Closed form for this series involving multiple binomial coefficients The series is:
$$\sum_{k=1}^m k {n-1 \choose n-k}{m \choose k}$$
where $m \leq n$. Is there a better form for this series? Perhaps, can we clean up the binomial coefficients somehow to make the series less complicated (or better yet, make it not a series anymore)?
I'm at a loss after trying examples for the last hour to try and get a better understanding. As for combinatorial arguments, I think ${n-1 \choose n-k}$ may be related to stars and bars, in particular with $n-k$ options and $k$ choices.
 A: This is same as
$$m \sum_{k=1}^m \binom{n-1}{n-k}\binom{m-1}{k-1} = m \sum_{t=0}^{m-1} \binom{n-1}{t}\binom{m-1}{t} = m \sum_{t=0}^{m-1} \binom{n-1}{t}\binom{m-1}{m-1-t}= m\binom{n+m-2}{m-1}.$$
The last step uses https://en.wikipedia.org/wiki/Vandermonde%27s_identity .
A: First, notice that $${n-1 \choose n-k} = {(n-k) + (k-1) \choose n-k}$$
so this term is stars and bars with $n-k$ items and $k$ options. Equivalently, it is the number of solutions to
$$x_1 + \dots + x_{k} = n$$
where each $x_i$ must be at least $1$.
Now, this is multiplied by ${m \choose k}$. This is the number of binary $m$-tuples with $k$ ones. For each $m$-tuple, consider all of the $m$-tuples formed by replacing the $i$-th appearance of $1$ with $x_i$ over all  solutions $(x_1, \dots, x_k)$ to the constraints above. For example, if we have $n = 5, k = 2, m=2$, a solution $x_1 = 3, x_2 = 2$ and a $3$-tuple $(0,1,1)$, then we consider the $3$-tuple $(0, 3, 2)$. 
Now, if we take ${n-1 \choose n-k}{m \choose k}$ and sum from $k=1$ to $m$, we are just finding the number of solutions to 
\begin{equation}
x_1 + \dots + x_m = n
\end{equation}
with each $x_i$ just non-negative. Denote the set of solutions by $S_n$. Essentially, the sum breaks this number up into groups based on the number of non-zero $x_i$'s. Finally, by multiplying by $k$ we are counting the number of non-zero $x_i$'s over all solutions. 
We can find this in a clever way: consider some  $(x_1, \dots, x_m) \in S_{n-1}$. By adding $1$ to each of the arguments one at a time, we create a symmetric relation on $S_{n-1} \times S_n$ (e.g. $(1, 2)$ is related to $(1, 3)$ and $(1, 2)$ is related to $(2, 2)$). We can ask, for $(x_1, \dots, x_m) \in S_n$, how many elements is it related to in $S_{n-1}$? It is not hard to see that this is the same as the number of non-zero $x_i$. Hence, it turns out that the sum in question is the same as the number of ordered pairs in this relation.
Well, each element of $S_{n-1}$ gives rise to $m$ ordered pairs, and $|S_{n-1}| = {n + m - 2 \choose n - 1}$, so finally the answer is
$$m {n + m - 2 \choose n-1}.$$
A: Here’s a purely combinatorial argument.
You have a pool of $m$ men and $n-1$ women from which to choose a team of $n$ players. You also have to choose one of the chosen players to be captain, and the outmoded rules under which you’re operating require that the captain be a man. Let $k$ be the number of men chosen; clearly $1\le k\le m$. There are $\binom{m}k$ ways to choose the $k$ men, $k$ ways to pick one of the chosen men to be captain, and $\binom{n-1}{n-k}$ ways to choose $n-k$ women to fill out the team of $n$ players. Thus, there are 
$$k\binom{n-1}{n-k}\binom{m}k$$
ways to pick a team with $k$ men and appoint one of the men captain. Summing over the possible values of $k$, we see that there are altogether
$$\sum_{k=1}^mk\binom{n-1}{n-k}\binom{m}k$$
ways to form a team and choose a captain.
Alternatively, we could begin by picking the captain; we can choose any of the men for this rôle, so there are $m$ ways to do so. That leaves a pool of $(n-1)+(m-1)=n+m-2$ players, from whom we can select any $n-1$ to complete the team; this can be done in $\binom{n+m-2}{n-1}$ ways. Thus, there are 
$$m\binom{n+m-2}{n-1}$$
ways to select the team and choose a captain, and we must have
$$\sum_{k=1}^mk\binom{n-1}{n-k}\binom{m}k=m\binom{n+m-2}{n-1}\;.$$
