Prove that $\sum_{k=0}^n (-1)^k\frac{{ {n}\choose{k}}{ {x}\choose{k}}}{{ {y}\choose{k}}} = \frac{{ {y-x}\choose{n}}}{{ {y}\choose{n}}}$ I need to prove this equation
$$\sum_{k=0}^n (-1)^k\frac{{ {n}\choose{k}}{ {x}\choose{k}}}{{ {y}\choose{k}}} = \frac{{ {y-x}\choose{n}}}{{ {y}\choose{n}}}$$
where $x$, $y$ and $n$ are nonnegative integers satisfying $y \geq n$.
The sign $(-1)^k$ suggests using binomial expansion, I've tried this, but without success. Is there a better way?

Standing the conditions you gave, the $y \choose n$ is not null, and we can multiply by it  both sides, giving
$$
\left( \begin{gathered}
  y - x \\ 
  n \\ 
\end{gathered}  \right) = \sum\limits_{0\, \leqslant \,k\, \leqslant \,n} {\left( { - 1} \right)^{\,k} \left( \begin{gathered}
  y \\ 
  n \\ 
\end{gathered}  \right)\left( \begin{gathered}
  n \\ 
  k \\ 
\end{gathered}  \right)\left( \begin{gathered}
  x \\ 
  k \\ 
\end{gathered}  \right)/\left( \begin{gathered}
  y \\ 
  k \\ 
\end{gathered}  \right)} 
$$
The RHS can be developed as

$$
\begin{gathered}
  \sum\limits_{0\, \leqslant \,k\, \leqslant \,n} {\left( { - 1} \right)^{\,k} \left( \begin{gathered}
  y \\ 
  n \\ 
\end{gathered}  \right)\left( \begin{gathered}
  n \\ 
  k \\ 
\end{gathered}  \right)\left( \begin{gathered}
  x \\ 
  k \\ 
\end{gathered}  \right)/\left( \begin{gathered}
  y \\ 
  k \\ 
\end{gathered}  \right)}  =  \hfill  \\
   = \sum\limits_{0\, \leqslant \,k\, \leqslant \,n} {\left( { - 1} \right)^{\,k} \left( \begin{gathered}
  y \\ 
  k \\ 
\end{gathered}  \right)\left( \begin{gathered}
  y - k \\ 
  n - k \\ 
\end{gathered}  \right)\left( \begin{gathered}
  x \\ 
  k \\ 
\end{gathered}  \right)/\left( \begin{gathered}
  y \\ 
  k \\ 
\end{gathered}  \right)}  =  \hfill  \\
   = \sum\limits_{0\, \leqslant \,k\, \leqslant \,n} {\left( { - 1} \right)^{\,k} \left( \begin{gathered}
  y - k \\ 
  n - k \\ 
\end{gathered}  \right)\left( \begin{gathered}
  x \\ 
  k \\ 
\end{gathered}  \right)}  =  \hfill   \\
   = \sum\limits_{0\, \leqslant \,k\, \leqslant \,n} {\left( \begin{gathered}
  y - k \\ 
  n - k \\ 
\end{gathered}  \right)\left( \begin{gathered}
  k - x - 1 \\ 
  k \\ 
\end{gathered}  \right)}  =  \hfill    \\
   = \left( \begin{gathered}
  y - x \\ 
  n \\ 
\end{gathered}  \right) \hfill \\ 
\end{gathered} 
$$
  which is valid for $x$ and $y$ even complex, with the only condition that ${y \choose n }\ne 0$.



*

*---Note ----*  


1st step) Trinomial Revision :
$\left( \begin{gathered}
  y \\ 
  n \\ 
\end{gathered}  \right)\left( \begin{gathered}
  n \\ 
  k \\ 
\end{gathered}  \right) = \left( \begin{gathered}
  y \\ 
  k \\ 
\end{gathered}  \right)\left( \begin{gathered}
  y - k \\ 
  n - k \\ 
\end{gathered}  \right)$
2nd step) simplifying $ {y \choose k }$
3rd step) VanderMonde Convolution
