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:



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> **QUESTION.** is there a combinatorial reason why the sum in (1) is independent of $\ell$?
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Is there a combinatorial reason for variable-independence of this binomial-coefficient identity?
*(I am late to post a very similar answer to the already given one, yet I'd like to post it as well, since it differs in some detail)*


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 $\{x\_0<x\_1<\dots<x\_t\}$ 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.


