$$\sum_{k=q+1}^{n-p+q+1} \binom{n-k}{p-q-1} \binom{k-1}{q} = \binom{n}{p}$$
I think this is some form of vandermondes identity,but the indexes are way different. Any hints?
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The left hand appears to be a function of $q$, in addition to $n,p$. But of course the right hand isn't. Can you prove that expression is independent of $q$? For instance, if $q+1>n$, isn't the left hand $0$? – lulu Jan 29 '25 at 12:50
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2I'm guessing $0\le q<p\le n$. You want to build a subset $X$ of ${1,2,\ldots,n}$ with $p$ elements. Let $k$ denote the $(q+1)$-th lowest element of $X$. Then $k$ belongs to ${q+1,\ldots,n-p+q+1}$. To construct such a subset $X$ with a given $k$, it suffices to choose the $q$ elements which must be strictly less than $k$ (so $\binom{k-1}q$ choices), and then choose the $p-q-1$ other elements which must be strictly above $k$ (so $\binom{n-k}{p-q-1}$ choices). – nejimban Jan 29 '25 at 14:11
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I agree with the comment of nejimban's set theory approach, that it is as if the position of the $~(q+1)$-th selected element is assigned the variable $~k,~$ and the value $~k~$ roams from its lower bound through its upper bound, inclusive. However, I suspect that when this type of combinatoric assertion permits a set theory proof, that the existence of a separate algebraic proof is (at least) plausible. I am curious what such an algebraic proof would look like. Personally, I am not qualified to attempt to derive such an algebraic proof. – user2661923 Jan 29 '25 at 14:52