# Is there a combinatorial reason for variable-independence of this binomial-coefficient identity?

Consider the following identity $$\sum_{n=0}^{R-t}\binom{n+\ell}n\binom{R-\ell-n}{R-t-n}=\binom{R+1}{t+1}.\tag1$$ It is relatively easy to give an algebraic or mechanical proof of (1). But, I like to ask:

QUESTION. is there a combinatorial reason why the sum in (1) is independent of $$\ell$$?

• This is ('negative') Chu-Vandermonde, as the LHS is $(-1)^{R-t} \sum_n \binom{-\ell -1}{n}\binom{\ell-t-1}{R-t-n} =(-1)^{R-t} \binom{-t-2}{R-t}=\binom{R+1}{t+1}$. Dec 3, 2020 at 17:35

Changing the summation index to $$m=n+\ell$$, the identity writes $$\sum_{m=\ell}^{R-t+\ell } {m\choose \ell}{R-m\choose t-\ell}={R+1\choose t+1}.$$
Given natural numbers $$\ell\le t\le R$$, and $$m$$, we may consider the class of those $$(t+1)$$-subsets $$\{x_0 of $$\{0,1,\dots,R\}$$ such that $$x_\ell=m$$: these are exactly $${m\choose \ell}{R-m\choose t-\ell}$$ (indeed the $$\ell$$ elements $$x_0,\dots, x_{\ell-1}$$ can be chosen freely into $$\{0,\dots, m-1\}$$, and so can the $$t-\ell$$ elements $$x_{\ell+1},\dots,x_t$$ into $$\{m+1,\dots,R\}$$. These classes, for $$\ell\le m\le R-t+\ell$$ form a partition of all $$(t+1)$$-subsets of $$[R+1]$$, whence the sum of their cardinality is independent of $$\ell$$ and the identity.
Well, here's a combinatorial proof of the identity anyway (but not a direct proof of independence of $$\ell$$).
Write $$N = R - t$$. Then the identity is $$\sum_{n=0}^N \binom{n + \ell}{n} \binom{R - n-\ell}{N - n} = \binom{R + 1}{N}.$$ The RHS is the number of $$N$$-sets $$\{x_1 < \cdots < x_N\} \subset \{1, \dots, R+1\}$$. Consider the last index $$n$$ such that $$x_n \leq n + \ell$$ (say $$n = 0$$ if there is no such index). Then $$\{x_1 < \cdots < x_n\} \subset \{1, \dots, n+\ell\}$$ and $$\{x_{n+1} < \cdots < x_N\} \subset \{n+\ell+1, \dots, R+1\}$$. The number of ways of making these choices is the LHS.