Simplifying a binomial expression I am interested in counting the number of hyperedges in the complete, $t$-uniform hypergraph on $n$ vertices which intersect a specified set of $k$ vertices.  This is trivial, the answer is:
$$\sum_{i=1}^t {k \choose i}{n-k \choose t-i}.$$
My questions is whether there is a nice simplification of this expression; I'd like to get rid of the sum if possible.  Anyone know?
Thanks a lot for the help!

We can use the Chu-Vandermonde identity (see Equation 7 in linked page):-
$$\sum_{i=0}^t {k \choose i}{n-k \choose t-i} = {n \choose t}$$
so that the sum can be simplified to
$$\sum_{i=1}^t {k \choose i}{n-k \choose t-i} = {n \choose t}-{n-k \choose t}$$
