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?  :/
 A: Using Negative Binomial Coefficients and Vandermonde's Identity, we get
$$
\begin{align}
\sum_{k=0}^n\binom{x+n-k-1}{n-k}\binom{y+k-1}{k}
&=\sum_{k=0}^n(-1)^{n-k}\binom{-x}{n-k}(-1)^k\binom{-y}{k}\tag{1}\\
&=(-1)^n\sum_{k=0}^n\binom{-x}{n-k}\binom{-y}{k}\tag{2}\\
&=(-1)^n\binom{-x-y}{n}\tag{3}\\
&=\binom{n+x+y-1}{n}\tag{4}
\end{align}
$$
Explanation:
$(1)$: Negative Binomial Coefficient conversion
$(2)$: algebra
$(3)$: Vandermonde's Identity
$(4)$: Negative Binomial Coefficient conversion
A: 
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 $t^n$.
A: We have $$\sum_{k=0}^n {y-1+k\choose k} {x-1+n-k\choose n-k} = \sum_{k=0}^n [z^{k}] \frac{1}{(1-z)^y} [z^{n-k}] \frac{1}{(1-z)^x} \\ = [z^n]  \frac{1}{(1-z)^y} \frac{1}{(1-z)^x} = [z^n] \frac{1}{(1-z)^{x+y}} ={x+y-1+n\choose n}.$$
A: $\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle}
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 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\half}{{1 \over 2}}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\iff}{\Longleftrightarrow}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\Li}[1]{\,\mathrm{Li}_{#1}}
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 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\ul}[1]{\underline{#1}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

Hereafter, I'll use the identities:
  $$
\left\lbrace\begin{array}{rcl}
\ds{a \choose b} & \ds{=} & \ds{{-a + b - 1 \choose b}\pars{-1}^{b}
\,,\quad b \in \mathbb{Z}}
\\[5mm]
\ds{a \choose b} & \ds{=} &
\ds{\oint_{\verts{z} = 1^{-}}{\pars{1 + z}^{a} \over z^{b + 1}}
\,{\dd z \over 2\pi\ic}\,,\quad b \in \mathbb{Z}}
\\[5mm]
\ds{\pars{1 + z}^{a}} & \ds{=} & \ds{\sum_{b = 0}^{\infty}{a \choose b}z^{b}\,,
\quad\verts{z} < 1}
\end{array}\right.
$$

\begin{align}
&\color{#f00}{\sum_{k = 0}^{n}{x + n - k - 1 \choose n - k}
{y + k - 1 \choose k}} =
\sum_{k = 0}^{\infty}\bracks{{-x \choose n - k}\pars{-1}^{n - k}}
\bracks{{-y \choose k}\pars{-1}^{k}}
\\[5mm] = &
\pars{-1}^{n}\sum_{k = 0}^{n}{-y \choose k}\oint_{\verts{z} = 1^{-}}
{\pars{1 + z}^{-x} \over z^{n - k + 1}}\,{\dd z \over 2\pi\ic} =
\pars{-1}^{n}\oint_{\verts{z} = 1^{-}}
{\pars{1 + z}^{-x} \over z^{n + 1}}\sum_{k = 0}^{\infty}{-y \choose k}z^{k}\,{\dd z \over 2\pi\ic}
\\[5mm] = &\
\pars{-1}^{n}\oint_{\verts{z} = 1^{-}}
{\pars{1 + z}^{-x - y} \over z^{n + 1}}\,{\dd z \over 2\pi\ic} =
\pars{-1}^{n}{-x - y \choose n} =
\pars{-1}^{n}{x  + y + n - 1 \choose n}\pars{-1}^{n}
\\[5mm] = &\
\color{#f00}{{x  + y + n - 1 \choose n}}
\end{align}
A: The multiset numbers $(\!\tbinom{m}{r}\!)=\binom{m+r-1}{r}$ count the multisets of cardinality $r$ with elements drawn from a set of size $m$. Using the multiset numbers, the identity becomes:
$$\left(\!\tbinom{x+y}{n}\!\right)=\sum_{k=0}^n \left(\!\tbinom{x}{n-k}\!\right) \left(\!\tbinom{y}{k}\!\right)$$
This has more or less the same combinatorial interpretation as the Vandermonde convolution identity for binomial coefficients. Any $n$-multiset built from the union of two sets of sizes $x$ and $y$ is uniquely a union of an $(n-k)$-multiset built from the set of size $x$ and a $k$-multiset built from the set of size $y$ for some $0\le k\le n$.
A: Geometric Solution:  Assume that the based field is $\mathbb{K}$, which is of characteristic $0$.  Prove that the equality holds when $x$ and $y$ are integers (for example, via combinatorial arguments).  Now, $\mathbb{Z}^2$ (as well as $\mathbb{N}^2$) is a Zariski-dense subset of $\mathbb{K}^2$.  Therefore, the equality holds for all $(x,y)\in\mathbb{K}^2$, noting that both sides are polynomials.
