<s>[INST] Proving that ${x +y+n- 1 \choose n}= \sum_{k=0}^n{x+n-k-1 \choose n-k}{y+k-1 \choose k} $ How can I prove that $${x +y+n- 1 \choose n}= \sum_{k=0}^n{x+n-k-1 \choose n-k}{y+k-1 \choose k} $$
I tried the following:
We use the falling factorial power: 
$$y^{\underline k}=\underbrace{y(y-1)(y-2)\ldots(y-k+1)}_{k\text{ factors}},$$
so that $\binom{y}k=\frac{y^{\underline k}}{k!} .$ 
Then
$${x +y+n- 1 \choose n} = \frac{(x +y+n- 1)!}{n! ((x +y+n- 1) - n)!} = 
\frac{1}{n!}. (x +y+n \color{#f00}{-1})^{\underline n} $$
And
$$ {x+n-k-1 \choose n-k}{y+k-1 \choose k}$$
$$\frac{1}{(n-k)!}.(x+n-k-1)^{\underline{n-k}}.\frac{1}{k!}.(y+k-1)^{\underline{k}}$$
$$\frac{1}{k!.(n-k)!}.(x+n-k-1)^{\underline{n-k}}.(y+k-1)^{\underline{k}}$$
$$\sum_{k=0}^n{n \choose k}(x+n-k-1)^{\underline{n-k}}.(y+k-1)^{\underline{k}}$$
According to the Binomial-coefficients:
$$ ((x+n-k-1) + (y+k-1))^{\underline{n}}$$
$$ (x+y+n\color{#f00}{- 2})^{\underline{n}}$$
What is wrong ? und How can I continue?  :/
 [/INST] 
Hint: The binomial formula with the Cauchy product
  \begin{align*}
(x+y)^n=\sum_{k=0}^n\binom{n}{k}x^ky^{n-k}
\end{align*}
  does not use falling factorials $x^{\underline{k}}$ resp. $y^{\underline{n-k}}$.

Here   is a step by step answer similar to that by @MarkoRiedel.  It's convenient to use the coefficient of operator $[z^k]$ to denote the coefficient of $z^k$ in a series. This way we can write e.g.
\begin{align*}
\binom{n}{k}=[z^k](1+z)^n
\end{align*}

We obtain 
\begin{align*}
\sum_{k=0}^{n}&\binom{x-1+n-k}{n-k}\binom{y-1+k}{k}\\
&=\sum_{k=0}^{\infty}\binom{x-1+n-k}{n-k}\binom{-y}{k}(-1)^k\tag{1}\\
&=\sum_{k=0}^\infty [t^{n-k}](1+t)^{x-1+n-k}[z^k](1+z)^{-y}(-1)^k\tag{2}\\
&=[t^n](1+t)^{x-1+n}\sum_{k=0}^\infty(-1)^kt^k(1+t)^{-k}[z^k](1+z)^{-y}\tag{3}\\
&=[t^n](1+t)^{x-1+n}\sum_{k=0}^\infty\left(-\frac{t}{1+t}\right)^k[z^k](1+z)^{-y}\\
&=[t^n](1+t)^{x-1+n}\left(1-\frac{t}{1+t}\right)^{-y}\tag{4}\\
&=[t^n](1+t)^{x+y-1+n}\tag{5}\\
&=\binom{x+y-1+n}{n}
\end{align*}
  and the claim follows.

Comment:


*

*In (1) we use  the binomial identity $\binom{-p}{q}(-1)^q=\binom{p+q-1}{q}$ and we extend the upper limit of the series to $\infty$ without changing anything since we are adding zeros only.

*In (2) we apply the coefficient of operator twice.

*In (3) we do some rearrangements by using the linearity of the coefficient of operator  and  we also  use  the  rule
\begin{align*}
[z^{p-q}]A(z)=[z^p]z^{q}A(z)
\end{align*}

*In (4) we apply the substitution rule
\begin{align*}
A(t)=\sum_{k=0}^\infty a_kt^k=\sum_{k=0}^\infty t^k[z^k]A(z)\\
\end{align*}
with $z=-\frac{t}{1+t}$.

*In (5) we do some simplifications.

*In (6) we select the coefficient from $t^n$.
</s>