It's known that $\binom{n}{r} = \binom{n}{s}$ if and only if $r = s$ or $r = n - s$.

If $n \neq m$, is it true that $\binom{n}{s} \binom{m}{r} = \binom{n}{k} \binom{m}{\ell}$ if and only if ($s = k$ or $s = n - k$) and ($r = \ell$ or $r = m - \ell$)?

  • $\begingroup$ Since the claim above is not true, here's the follow-up question: If $n \neq m$ and $r,s,k,\ell > 0$, when is $\binom{n}{s} \binom{m}{r} = \binom{n}{k} \binom{m}{\ell}$? $\endgroup$ – Nate Gallup Sep 2 '15 at 19:20


  • $\begingroup$ How did you find this? (It's a consequence of $4!5!=2!2!6!$.) $\endgroup$ – Akiva Weinberger Sep 2 '15 at 19:45
  • 3
    $\begingroup$ @columbus8myhw: It immediately occurred to me that $\binom52=2\binom51$, so I did a quite mental run-through for another pair of binomial coefficients with the same upper number, one of which was twice the other. $\endgroup$ – Brian M. Scott Sep 2 '15 at 19:51

I don't have any good ideas to answer the follow up question, but it's worth noting there are infinitely many solutions to that equation outside of the two possibilities mentioned in the original question. For example, if $m=\binom{n}{k}$ for some $1<k<n-1$, then $\binom{n}{k}\binom{m}{0}=m=\binom{n}{0}\binom{m}{1}$.

EDIT: Okay, so the OP wants nontrivial answers, eh? Well, just generalize Brian's answer. Why does it work? Well, we have the identity $$\binom{n}{k}=\binom{n}{k-1}\frac{n-k+1}{k}$$ so one way to transmute a product into binomials into another such product is $$\binom{n}{k}\binom{m}{\ell}=\binom{n}{k-1}\binom{m}{\ell+1}\frac{(n-k+1)(\ell+1)}{(m-\ell)k}.$$ If we can come up with $m\neq n$, $1<\ell<m-1$, and $2<k<n$ such that $(\ell+1)(n-k+1)=(m-\ell)k$ then we get examples. But these are more or less independent parameters, just pick stuff and see what works. For example, suppose we always want $k=\ell=2$. Then we need $3(n-1)=2(m-2)$ hence $2m=3n+1$. This only has solutions when $n$ is odd, and we get: $$n=1, m=2: \text{trivial case (both sides 0)}$$ $$n=3, m=5: \text{trivial case since } k-1=m-k$$ $$n=5, m=8: \text{Brian's example}$$ $$n=7, m=11: \binom{7}{2}\binom{11}{2}=165=\binom{7}{1}\binom{11}{3}$$ It's not hard to see that from here on out we get new examples. If we write $n=2r+1$, we see that we will always have $\binom{2r+1}{2}\binom{6r+2}{2}=\binom{2r+1}{1}\binom{6r+2}{3}$, so there is a slew of examples. We could obviously cook up more by varying $k$ and $\ell$. (e.g. $\ell=k=3$ yields $\binom{3r-1}{3}\binom{4r-1}{3}=\binom{3r-1}{2}\binom{4r-1}{4}$, so $\binom{8}{3}\binom{11}{3}=\binom{8}{2}\binom{11}{4}$)

I have no idea at the moment how to come up with a general solution though.

  • $\begingroup$ Good point, thanks! I should have mentioned that I want to require $r,s,k,\ell$ to be positive. I've edited the question to reflect that, sorry! $\endgroup$ – Nate Gallup Sep 2 '15 at 19:19

This is a comment, not a solution, but recall the identity $(*)$ $\binom{n}{m} {m \choose k}={n \choose k}{n-k \choose m-k}$ and letting $n=8, m=5$ we can see that Brian's counterexample to your original claim is of this form as $k=3$ and since ${5 \choose 2}={5 \choose 3}$, $k=2$ is equivalent, thus we may interchange the products on either resulting $rhs$ of $(*)$ justified in this case by the fact that ${8 \choose 3}=2{8 \choose 2}$ as well as ${8-3 \choose 5-3}=\frac{1}{2}{8-2 \choose 5-2}$. So this would seem to suggest that $m=n-k_i$ for some $k_i$ such that for all $i,j$, ${m \choose k_i}={m \choose k_j}$. It would appear possible that all the solutions to your problem must be equivalent to an expression of the form $(*)$ by some series of relations gleaned from identities as previously mentioned, which seem to extend the conditions of your original claim, that is any solution to your problem that doesn't satisfy the conditions of your original claim must be equivalent to one that in some sense does. Just a thought though, and it could be a useless one, my apologies if thats the case.


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