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!
 A: 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
$$\sum_{i=1}^t {k \choose i}{n-k \choose t-i} = {n \choose t}-{n-k \choose t}$$
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
 \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack}
 \newcommand{\dd}{{\rm d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\dsc}[1]{\displaystyle{\color{red}{#1}}}
 \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}
 \newcommand{\half}{{1 \over 2}}
 \newcommand{\ic}{{\rm i}}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\Li}[1]{\,{\rm Li}_{#1}}
 \newcommand{\pars}[1]{\left(\, #1 \,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,}
 \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$
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,
$$
\color{#66f}{%
\sum_{j = 1}^{t}\pars{k \atop j}\pars{n - k \atop t - j}}
=\color{#66f}{\pars{n \atop t} - \pars{n - k \atop t}}
$$
