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?
 A: Note that
$$\binom{y}n\binom{n}k=\binom{y}k\binom{y-k}{n-k}\;,$$
so
$$\binom{y}n\cdot\frac{\binom{n}k\binom{x}k}{\binom{y}k}=\binom{y-k}{n-k}\binom{x}k\;,$$
and after multiplication by $\binom{y}n$, the proposed identity reduces to
$$\sum_{k=0}^n(-1)^k\binom{x}k\binom{y-k}{n-k}=\binom{y-x}n\;.\tag{1}$$
$\binom{y-x}n$ is the number of $n$-element subsets of $[y]\setminus[x]$, where as usual $[x]=\{1,\ldots,x\}$. For $k\in[x]$ let $\mathscr{A}_k$ be the family of $n$-element subsets of $[y]$ that contain $k$. Then
$$\left|\,\bigcap_{k\in I}\mathscr{A}_k\,\right|=\binom{y-|I|}{n-|I|}$$
whenever $\varnothing\ne I\subseteq[x]$, so by the inclusion-exclusion principle we have
$$\begin{align*}
\left|\,\bigcup_{k\in[x]}\mathscr{A}_k\,\right|&=\sum_{\varnothing\ne I\subseteq[x]}(-1)^{|I|+1}\binom{y-|I|}{n-|I|}\\
&=\sum_{k=1}^x(-1)^{k+1}\binom{x}k\binom{y-k}{n-k}\\
&=\sum_{k=1}^n(-1)^{k+1}\binom{x}k\binom{y-k}{n-k}\;,
\end{align*}$$
assuming that $x\ge n$. (Recall that by definition $\binom{y-k}{n-k}=0$ when $n-k<0$.) This is the number of $n$-element subsets of $[y]$ that do intersect $[x]$, so 
$$\begin{align*}\binom{y-x}n&=\binom{y}n-\left|\,\bigcup_{k\in[x]}\mathscr{A}_k\,\right|\\
&=\binom{y}n-\sum_{k=1}^n(-1)^{k+1}\binom{x}k\binom{y-k}{n-k}\\
&=\binom{y}n+\sum_{k=1}^n(-1)^k\binom{x}k\binom{y-k}{n-k}\\
&=\sum_{k=0}^n(-1)^k\binom{x}k\binom{y-k}{n-k}\;,
\end{align*}$$
as desired.
Added: as darij grinberg pointed out in the comments, the combinatorial interpretation requires the assumption that $y\ge x$. For each $x$ the combinatorial argument establishes $(1)$ for each integer $y\ge x$ and each side of $(1)$ is an $n$-th degree polynomial in $y$, so $(1)$ must be a polynomial identity.
A: 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
A: Here is a variation based upon Vandermonde's identity. It is convenient to simplify OPs identity 
\begin{align*}
\sum_{k=0}^n (-1)^k\frac{{ {n}\choose{k}}{ {x}\choose{k}}}{{ {y}\choose{k}}} = \frac{{ {y-x}\choose{n}}}{{ {y}\choose{n}}}
\end{align*}
by multiplying both sides with $\binom{y}{n}$ and

we claim
\begin{align*}
\sum_{k=0}^n(-1)^k\binom{x}{k}\binom{y-k}{n-k}=\binom{y-x}{n}
\end{align*}
We obtain
  \begin{align*}
\sum_{k=0}^n(-1)^k\binom{x}{k}\binom{y-k}{n-k}&=(-1)^n\sum_{k=0}^n\binom{x}{k}\binom{n-y-1}{n-k}\tag{1}\\
&=(-1)^n\binom{x+n-y-1}{n}\tag{2}\\
&=\binom{y-x}{n}\tag{3}
\end{align*}
  and the claim follows.

Comment:


*

*In (1) we apply to $\binom{y-k}{n-k}$ the binomial identity $\binom{-p}{q}=\binom{p+q-1}{q}(-1)^q$

*In (2) we apply Vandermonde's identity to the right-hand series in (1)

*In (3) we  apply  the binomial identity from (1) to $\binom{x+n-y-1}{n}$
