Update: I have dramatically improved my original proof, and left it below a horizontal rule._

I will prove the following stronger result from A.K. Gupta, Generalized hidden hexagon squares, Fibonacci Quarterly, 1974: $$ \binom{n}{m-r} \binom{n-s}{m} \binom{n+r}{m+s} = \binom{n-s}{m-r} \binom{n}{m+s} \binom{n+r}{m} $$

Proof:

Consider the following problem: divide a set of size $3(a+b) + p + q$ into sets of size $a+p$, $a+q$, $b-p$, $b-q$, $b$, and $a+p+q$.

You can lump them as $(a+p)+(b-p)$, $(a+q)+(b)$, and $(a+p+q)+(b-q)$, whence the problem becomes into two parts: first divide the big set into sizes $a+b$, $a+b+q$, and $a+b+p$, and then do $\binom{a+b}{b-p} \binom{a+b+q}{b} \binom{a+b+p}{b-q}$ choices.

Alternately, you can switch the roles of $p$ and $q$ in the previous paragraph; either way you first divide the same big set into the same three pieces, but then how you divvy up the pieces looks different. Note that if all six of the small sets have nonnegative size, then so do the three big sets. So we have proven: $$ \binom{a+b}{b-p} \binom{a+b+q}{b} \binom{a+b+p}{b-q} = (p\leftrightarrow q) $$ with no conditions on the signs of any number (only that $a,b,p,q$ are all integers).

Setting $n=a+b$, $m=b$, $r=p$, $s=-q$ (for example) gives the desired result.

My earlier proof; some comments apply to it:

Let me write $\binom{a}{b,c}$ for the number of ways to choose two disjoint sets, of sizes $b,c$ respectively, from a set of size $a$. There are a few ways to do this: you could first pick $b$ of them, and then pick $c$ from the remaining, or you could pick $b+c$, and then decide how to divy them up, or... Thus, we have the following equalities: $$ \binom{n}{m-r,s} = \binom{n}{m-r}\binom{n-(m-r)}{s} = \binom{n-s}{m-r} \binom{n}{s} $$ $$ \binom{n}{m,s} = \binom{n-s}{m} \binom{n}{s} = \binom{n}{m+s} \binom{m+s}{s} $$ $$ \binom{n+r}{m,s} = \binom{n+r}{m+s} \binom{m+s}{s} = \binom{n+r}{m} \binom{(n+r)-m}{s} $$ Multiplying these equations together and canceling like terms gives the desired equality.

This does not provide a combinatorial interpretation when both sides are $0$ --- indeed, it applies only when all of $s$, $m$, $n-s$, and $n+r-m-s$ are nonnegative integers, as in other situations I had asked to divide by $0$.

I will not try to improve this, as you asked for a combinatorial proof, and I won't try to give combinatorial interpretation to "the number of ways to pick negatively many objects form a set with negative size".

In words, I start with sets of size $n$, $n$, and $n+r$, and remove a set of size $s$ from each, and then remove from what remains sets of size $m-r$, $m$, and $m$; I count the ways to do this in various ways, and notice one coincidence, that $n-(m-r) = (n+r) - m$.

Update: I have dramatically improved my original proof, and left it below a horizontal rule._

I will prove the following stronger result from A.K. Gupta, Generalized hidden hexagon squares, Fibonacci Quarterly, 1974: $$ \binom{n}{m-r} \binom{n-s}{m} \binom{n+r}{m+s} = \binom{n-s}{m-r} \binom{n}{m+s} \binom{n+r}{m} $$

Proof:

Consider the following problem: divide a set of size $3(a+b) + p + q$ into sets of size $a+p$, $a+q$, $b-p$, $b-q$, $b$, and $a+p+q$.

You can lump them as $(a+p)+(b-p)$, $(a+q)+(b)$, and $(a+p+q)+(b-q)$, whence the problem becomes into two parts: first divide the big set into sizes $a+b$, $a+b+q$, and $a+b+p$, and then do $\binom{a+b}{b-p} \binom{a+b+q}{b} \binom{a+b+p}{b-q}$ choices.

Alternately, you can switch the roles of $p$ and $q$ in the previous paragraph; either way you first divide the same big set into the same three pieces, but then how you divvy up the pieces looks different. Note that if all six of the small sets have nonnegative size, then so do the three big sets. So we have proven: $$ \binom{a+b}{b-p} \binom{a+b+q}{b} \binom{a+b+p}{b-q} = (p\leftrightarrow q) $$ with no conditions on the signs of any number (only that $a,b,p,q$ are all integers).

Setting $n=a+b$, $m=b$, $r=p$, $s=-q$ (for example) gives the desired result.

My earlier proof; some comments apply to it:

Let me write $\binom{a}{b,c}$ for the number of ways to choose two disjoint sets, of sizes $b,c$ respectively, from a set of size $a$. There are a few ways to do this: you could first pick $b$ of them, and then pick $c$ from the remaining, or you could pick $b+c$, and then decide how to divy them up, or... Thus, we have the following equalities: $$ \binom{n}{m-r,s} = \binom{n}{m-r}\binom{n-(m-r)}{s} = \binom{n-s}{m-r} \binom{n}{s} $$ $$ \binom{n}{m,s} = \binom{n-s}{m} \binom{n}{s} = \binom{n}{m+s} \binom{m+s}{s} $$ $$ \binom{n+r}{m,s} = \binom{n+r}{m+s} \binom{m+s}{s} = \binom{n+r}{m} \binom{(n+r)-m}{s} $$ Multiplying these equations together and canceling like terms gives the desired equality.

This does not provide a combinatorial interpretation when both sides are $0$ --- indeed, it applies only when all of $s$, $m$, $n-s$, and $n+r-m-s$ are nonnegative integers, as in other situations I had asked to divide by $0$.

I will not try to improve this, as you asked for a combinatorial proof, and I won't try to give combinatorial interpretation to "the number of ways to pick negatively many objects form a set with negative size".

In words, I start with sets of size $n$, $n$, and $n+r$, and remove a set of size $s$ from each, and then remove from what remains sets of size $m-r$, $m$, and $m$; I count the ways to do this in various ways, and notice one coincidence, that $n-(m-r) = (n+r) - m$.

I will prove the following stronger result from A.K. Gupta, Generalized hidden hexagon squares, Fibonacci Quarterly, 1974: $$ \binom{n}{m-r} \binom{n-s}{m} \binom{n+r}{m+s} = \binom{n-s}{m-r} \binom{n}{m+s} \binom{n+r}{m} $$

Proof:

Let me write $\binom{a}{b,c}$ for the number of ways to choose two disjoint sets, of sizes $b,c$ respectively, from a set of size $a$. There are a few ways to do this: you could first pick $b$ of them, and then pick $c$ from the remaining, or you could pick $b+c$, and then decide how to divy them up, or... Thus, we have the following equalities: $$ \binom{n}{m-r,s} = \binom{n}{m-r}\binom{n-(m-r)}{s} = \binom{n-s}{m-r} \binom{n}{s} $$ $$ \binom{n}{m,s} = \binom{n-s}{m} \binom{n}{s} = \binom{n}{m+s} \binom{m+s}{s} $$ $$ \binom{n+r}{m,s} = \binom{n+r}{m+s} \binom{m+s}{s} = \binom{n+r}{m} \binom{(n+r)-m}{s} $$ Multiplying these equations together and canceling like terms gives the desired equality.

This does not provide a combinatorial interpretation when both sides are $0$ --- indeed, it applies only when all of $s$, $m$, $n-s$, and $n+r-m-s$ are nonnegative integers, as in other situations I had asked to divide by $0$.

I will not try to improve this, as you asked for a combinatorial proof, and I won't try to give combinatorial interpretation to "the number of ways to pick negatively many objects form a set with negative size".

In words, I start with sets of size $n$, $n$, and $n+r$, and remove a set of size $s$ from each, and then remove from what remains sets of size $m-r$, $m$, and $m$; I count the ways to do this in various ways, and notice one coincidence, that $n-(m-r) = (n+r) - m$.

Update: I have dramatically improved my original proof, and left it below a horizontal rule._

I will prove the following stronger result from A.K. Gupta, Generalized hidden hexagon squares, Fibonacci Quarterly, 1974: $$ \binom{n}{m-r} \binom{n-s}{m} \binom{n+r}{m+s} = \binom{n-s}{m-r} \binom{n}{m+s} \binom{n+r}{m} $$

Proof:

Consider the following problem: divide a set of size $3(a+b) + p + q$ into sets of size $a+p$, $a+q$, $b-p$, $b-q$, $b$, and $a+p+q$.

You can lump them as $(a+p)+(b-p)$, $(a+q)+(b)$, and $(a+p+q)+(b-q)$, whence the problem becomes into two parts: first divide the big set into sizes $a+b$, $a+b+q$, and $a+b+p$, and then do $\binom{a+b}{b-p} \binom{a+b+q}{b} \binom{a+b+p}{b-q}$ choices.

Alternately, you can switch the roles of $p$ and $q$ in the previous paragraph; either way you first divide the same big set into the same three pieces, but then how you divvy up the pieces looks different. Note that if all six of the small sets have nonnegative size, then so do the three big sets. So we have proven: $$ \binom{a+b}{b-p} \binom{a+b+q}{b} \binom{a+b+p}{b-q} = (p\leftrightarrow q) $$ with no conditions on the signs of any number (only that $a,b,p,q$ are all integers).

Setting $n=a+b$, $m=b$, $r=p$, $s=-q$ (for example) gives the desired result.

My earlier proof; some comments apply to it:

Let me write $\binom{a}{b,c}$ for the number of ways to choose two disjoint sets, of sizes $b,c$ respectively, from a set of size $a$. There are a few ways to do this: you could first pick $b$ of them, and then pick $c$ from the remaining, or you could pick $b+c$, and then decide how to divy them up, or... Thus, we have the following equalities: $$ \binom{n}{m-r,s} = \binom{n}{m-r}\binom{n-(m-r)}{s} = \binom{n-s}{m-r} \binom{n}{s} $$ $$ \binom{n}{m,s} = \binom{n-s}{m} \binom{n}{s} = \binom{n}{m+s} \binom{m+s}{s} $$ $$ \binom{n+r}{m,s} = \binom{n+r}{m+s} \binom{m+s}{s} = \binom{n+r}{m} \binom{(n+r)-m}{s} $$ Multiplying these equations together and canceling like terms gives the desired equality.

This does not provide a combinatorial interpretation when both sides are $0$ --- indeed, it applies only when all of $s$, $m$, $n-s$, and $n+r-m-s$ are nonnegative integers, as in other situations I had asked to divide by $0$.

I will not try to improve this, as you asked for a combinatorial proof, and I won't try to give combinatorial interpretation to "the number of ways to pick negatively many objects form a set with negative size".

In words, I start with sets of size $n$, $n$, and $n+r$, and remove a set of size $s$ from each, and then remove from what remains sets of size $m-r$, $m$, and $m$; I count the ways to do this in various ways, and notice one coincidence, that $n-(m-r) = (n+r) - m$.