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Usually in combinatorics, I love proofs by double counting. It gives me a very happy feeling to know a double counting proof. I feel I understand the problem better. A close younger sibling of this technique is to interpret a given expression as a solution to a smartly constructed counting problem.

So whenever somebody asks me to prove that a ratio involving factorials is an integer, I try to interpret the ratio as a solution to a counting problem. But throughout my counting life, I have encountered certain expressions which never admit an interpretative proof. One of them is jasoncube's question posted here.

I searched online and I could not find a slick proof for jasoncube's question. So this post has the following two questions:

1) For $m,n \in \mathbb{N}$ can you find a counting problem whose solution is $\dfrac{(2m)! (2n)!}{m! n! (m+n)!}$?

2) Is there any literature on the extent of this technique or it's limitations? That is, has anybody proved impossibility results for certain expressions?

Thanks,
Iso

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Isomorphism
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  • $\frac{(2 m)! (2 n)!}{m! n! (m + n)!}$ is the ratio of the number of ways of dividing $2 m + 2 n$ things into boxes of $m$, $n$, and $m + n$ things, vs the number of ways of dividing $2 m + 2 n$ things into boxes of $2 m$ and $2 n$ things. This is obviously a positive integer because given boxes of $2 m$ and $2 n$ things, just pick $m$ things from the first box and $n$ things from the second box and put the remainder into a third box. – Zhen Lin Dec 15 '12 at 21:03
  • @ZhenLin: The numerator for your counting problem is $(2m+2n)!$, but in my problem it is $(2m)!(2n)!$. – Isomorphism Dec 15 '12 at 21:04
  • Nope. The problem I describe is just the observation that $$\frac{(2 m)! (2 n)!}{m! n! (m + n)!} = \left. \frac{(2 m)!}{m! m!} \frac{(2 n)!}{n! n!} \middle/ \frac{(m + n)!}{m! n!} \right.$$ – Zhen Lin Dec 15 '12 at 21:08
  • I get $$\frac{(2m)!(2n)!}{m!n!(m+n)!}=\frac{\binom{2m}{m}\binom{2n}{n}}{\binom{m+n}{m}}$$ Not quite sure how @Zhen's interpretation arises. – hmakholm left over Monica Dec 15 '12 at 21:08
  • @ZhenLin: But what does that have to do with (say) "dividing $2m+2n$ things into boxes of $2m$ and $2n$ things"? For that there'd need to be a $\binom{2m+2n}{2m}$ somewhere, it seems. – hmakholm left over Monica Dec 15 '12 at 21:09
  • Sure. $$\frac{(2 m)! (2 n)!}{(m! n! (m + n)!} = \left. \frac{(2 m + 2 n)!}{m! n! (m + n)!} \middle/ \frac{(2 m + 2 n)!}{(2 m)! (2 n)!} \right.$$ – Zhen Lin Dec 15 '12 at 21:11
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    @ZhenLin: Okay. But I still don't get your argument that it is "obviously an integer" -- a partition into $m+n+(m+n)$ can arise from several different $2m+2n$ partitions using the procedure you describe, and vice versa. – hmakholm left over Monica Dec 15 '12 at 21:15
  • Indeed. That is why we have to divide by $\frac{(m + n)!}{m! n!}$. – Zhen Lin Dec 15 '12 at 21:15
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    @ZhenLin: No, I don't get it. You seem to be arguing that $\binom{2m+2n}{m,;n,;(m+n)}$ is "obviously" an integer multiple of $\binom{2m+2n}{2m}$, but that is not at all obvious to me. – hmakholm left over Monica Dec 15 '12 at 21:18
  • Obviously, $m + n + m + n$ is the same thing as a partition into $m + m + n + n$. Now regroup: a partition into $m + n + (m + n)$ arises from a partition into $m + n + m + n$ in $\frac{(m + n)!}{m! n!}$ different ways, and a partition into $2 m + 2 n$ arises from a partition into $m + m + n + n$ in $\frac{(2 m)!}{m! m!} \frac{(2 n)!}{n! n!}$ different ways. – Zhen Lin Dec 15 '12 at 21:29
  • @Zhen: (Typed before seeing your comment just above) Hmm, is this what you're getting at: $$ \binom{2n+2m}{n,;m,;(m+n)}\binom{m+n}{m,;n} = \binom{2n+2m}{n,;n,;m,;m} = \binom{2n+2m}{2n,;2m}\binom{2n}{n}\binom{2m}{m}$$ hence $$\frac{\binom{2n+2m}{n,;m,;(m+n)}}{\binom{2n+2m}{2n,;2m}}=\frac{\binom{2n}{n}\binom{2m}{m}}{\binom{m+n}{m,;n}}$$ That doesn't make it clearer to me that the common ratio is an integer. – hmakholm left over Monica Dec 15 '12 at 21:31
  • I don't really know how to explain the intuition. The point is that these combinatorial objects are related to each other in a completely homogeneous way, so the division is only to compensate for overcounting and will not produce a fractional number. It's not rigorous, but then again I am not a combinatorialist. – Zhen Lin Dec 15 '12 at 21:37
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    @Zhen: I think your intuition would be correct if there were overcounting only in one direction; that is, if your argument exhibited the ratio as the ratio of the cardinalities of two sets $A$, $B$ such that one element of $A$ corresponds to $r$ elements of $B$ but every element of $B$ only corresponds to one element of $A$; then the ratio would be the integer $r$. That's not the case here, however; every element of $B$ also corresponds to $s$ elements of $A$, and then the ratio is the ratio $r/s$ of two integers, and it remains to be explained why that ratio is itself an integer. – joriki Dec 15 '12 at 22:29

1 Answers1

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According to this question on Math Overflow ("Recursions which define polynomials") there is no known combinatorial interpretation of the numbers $$A(m,n) = \frac{(2m)! (2n)!}{m! n! (m+n)!}.$$

The question does link to Ira Gessel's paper "Super Ballot Numbers" (Journal of Symbolic Computation 14 (1992) 179--194). In Section 6 Gessel calls these "super Catalan numbers" and gives a few proofs of their integrality. Equation (32) consists of the formula $$\sum_n 2^{p-2n} \binom{p}{2n} A(m,n) = A(m,m+p), \:\:\: p \geq 0.$$ Gessel says that this formula, together with $A(0,0) = 1$ and $A(m,n) = A(n,m)$, "in principle... gives a combinatorial interpretation to $A(m,n)$, although it remains to be seen whether [the formula] can be interpreted in a 'natural' way."

So "no known combinatorial interpretation, but a recursive formula that might lead to one" appears to be the state of things at this point.

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Mike Spivey
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    The number is also a solution to the recursion: $B(m,n) = 4 B(m,n-1) - B(m+1,n-1)$ where $B(m,m) = \binom{2m}{m}$ according to the IMO 1972 solution at http://www.cs.cornell.edu/~asdas/imo/imo/isoln/isoln723.html – Isomorphism Dec 15 '12 at 22:38
  • @Isomorphism: That recurrence is at the end of Section 6 of Gessel's paper as well. – Mike Spivey Dec 15 '12 at 22:41
  • I had not seen the link to the paper. I am reading it now, looks wonderful! Thank you very much :))

    P.S: do you know any literature on the impossibility of proving a number cannot be obtained as a counting problem (I don't mind if the counting methods are restricted, I want to learn proof techniques). Thanks!

     – Isomorphism Dec 15 '12 at 22:45
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    @Isomorphism: I've never heard of anyone proving that a certain identity has no combinatorial proof. In fact, I would be surprised if someone did so. I cannot imagine how such a proof (of the non-existence of a combinatorial proof) would be constructed. But then proofs about proofs is getting far from my area of expertise. – Mike Spivey Dec 15 '12 at 22:53