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!

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We'll use the integral representation
$\ds{\pars{a \atop b}=\oint_{\verts{z} = R}\frac{\pars{1 + z}^{a}}{z^{b + 1}}\,
\frac{\dd z}{2\pi\ic}\,,\quad R > 0}$.

\begin{align}&\color{#66f}{%
\sum_{j = 1}^{t}\pars{k \atop j}\pars{n - k \atop t - j}}
=-\pars{n - k \atop t}
+\sum_{j = 0}^{\infty}\pars{k \atop j}\pars{n - k \atop t - j}
\\[5mm]&=-\pars{n - k \atop t} + \sum_{j = 0}^{\infty}\pars{k \atop j}\
\overbrace{\oint_{\verts{z}=1}\frac{\pars{1 + z}^{n - k}}{z^{t - j + 1}}
\,\frac{\dd z}{2\pi\ic}}^{\dsc{\pars{n - k \atop t - j}}}
\\[5mm]&=-\pars{n - k \atop t}
+\oint_{\verts{z}=1}\frac{\pars{1 + z}^{n - k}}{z^{t+ 1}}\ \overbrace{%
\sum_{j = 0}^{\infty}\pars{k \atop j}z^{j}}^{\dsc{\pars{1 + z}^{k}}}\
\,\frac{\dd z}{2\pi\ic}
=-\pars{n - k \atop t}\ +\ \overbrace{%
\oint_{\verts{z}=1}\frac{\pars{1 + z}^{n}}{z^{t+ 1}}\,\frac{\dd z}{2\pi\ic}}
^{\dsc{\pars{n \atop t}}}
\end{align}

Then,
$$
\color{#66f}{%
\sum_{j = 1}^{t}\pars{k \atop j}\pars{n - k \atop t - j}}
=\color{#66f}{\pars{n \atop t} - \pars{n - k \atop t}}
$$
