Looking for combinatorial identity: $\sum\limits_{j=0}^k{n \choose k-j}{m \choose j}$ Is there a nicer closed form expression for the following expression? $$\sum_{j=0}^k{n \choose k-j}{m \choose j}$$

Yes, it is known as Vandermonde's identity: $$\sum_{j=0}^{k}\binom{n}{k-j}\binom{m}{j}=\binom{n+m}{k}$$
The right side is the number of ways of picking $k$ elements from $m+n$ elements, while the left side is the number of ways of doing the same thing: first pick all $k$ from the $n$ elements and nothing from $m$, then $k-1$ from $n$ and $1$ from $m$, and so on till you pick none from the $n$ elements and all $k$ from the $m$ elements.
