I am trying to find specifically a combinatorial proof of the identity $$ (x+y)_{(n)} = \sum_k \binom n k x_{(k)}y_{(n-k)} $$ for all real $x,y$.
(The subscripted parentheses indicate the falling factorial.)
I've tried writing out both sides. Clearly there is supposed to be an analogy to the binomial theorem. And it seems pretty clear that somehow we ought to be able to choose $k$ of the $n$ factors in order to construct $x_{(k)}$ and then from the remaining factors construct $y_{(n-k)}$.
In the proof of the binomial theorem the construction is obvious: In the selected $k$ factors you choose $x$ for distribution, and from the remaining $n-k$ factors you choose $y$ for distribution. That exact same thing can't be done here because in the choice of $k$ arbitrary factors $(x+y-i_1)(x+y-i_2)\cdots (x+y-i_k)$ it's not clear how to construct $x_{(k)}$. I'm tempted to choose $x$ from the first factor, choose $x-1$ from the second, and so on. But what is left over for distribution for $y$ does not clearly yield $y_{(n-k)}$.
I've also thought about how perhaps we try to look for choices that can construct $x_{(k)}$. If we always started by taking $x$ from $(x+y)$ and $x-1$ from $(x+y-1)$ and so on $k$ times, then what would be left for distribution for $y$ would be $y^{n-k}$ if we were systematically just always grouping $x-i$ in the distribution. But this clearly doesn't lead to what we want. It also doesn't show how there would be any real act of choosing any $k$ things.
Consider a set $S$ partitioned as $S=X\sqcup Y$, such that $|X|=x$ and $|Y|=y$. Suppose we have to choose $n$ variables $v_1, \cdots, v_n$ taking values in $S$, such that they are mutually distinct, i.e., $v_i\ne v_j$ for $i\ne j$. This can be done in $(|S|)_{(n)}$ ways. Another way of choosing these $n$ variables, is to choose some of the values from $X$ and then the rest from $Y$. So for some $k\le n$, we first have to choose $k$ indices among the $n$ indices, which can be done in $\binom{n}{k}$ ways. Now if the index set $I=\{i_1, \cdots, i_k\}$ is chosen choose $v_j$ from $X$ if $j\in I$, or from $Y$ otherwise. Note that since $X$ and $Y$ are disjoint we should already have $v_j\ne v_k$, for $j\in I, k\not\in I$.