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I'm trying to prove the following question from IMO 1972

Prove that for any positive integers $m,n$, $$\frac{(2m)!(2n)!}{m!n!(m+n)!}$$ is always an integer.

I tried thinking of a combinatorial argument, and came up with this: Imagine that there are two bags with $2m$ and $2n$ objects respectively. This expression is the number of ways of selecting $m$ objects from the first bag and $n$ objects from the second bag, such that the order in which you select these objects is not important. For instance, you can keep switching between bags, and also changing the order in which you select the objects.

However, I am not convinced that this is a correct argument. Where am I going wrong? Thanks

  • how does your argument explain the (m+n)! in the denominator – Rohan Shinde Feb 07 '18 at 16:27
  • @Manthanein- That is what makes selecting the same set of objects in different orders equivalent. Without the $(m+n)!$, this would just mean you're selecting m objects form the first bag in a particular order, and then $n$ objects from the second bag in a particular order. –  Feb 07 '18 at 16:42
  • @AyushKhaitan with you approach what you get is really ${2m\choose m} {2n\choose n} $. More precisely, you can't "keep switching between bags" as you say, because that would make the numerator $(2m+2n)!$ – Arnaud Mortier Feb 07 '18 at 16:46
  • @ArnaudMortier- I don't think that's true. I'm not choosing $m+n$ objects from $2m+2n$ objects. There's the additional constraint that I'm choosing $m$ objects form the given $2m$ objects and $n$ objects from the given $2n$ objects. –  Feb 07 '18 at 16:49
  • @AyushKhaitan I know that this is what you want to do, but in this case you get ${2m\choose m}{2n\choose n}$. – Arnaud Mortier Feb 07 '18 at 16:52
  • @ArnaudMortier- Yes, but ${2m \choose m} {2n\choose n}$ allows us to choose $m$ objects from the first bag, and $n$ objects from the second bag, but doesn't allow us to switch bags at will. I figured that dividing by $(m+n)!$ instead of the usual $m!n!$ would allow us also to switch the bags at will. –  Feb 07 '18 at 17:01
  • @AyushKhaitan If you can switch bags at will, the numerator will not be the same, it will be (2n+2m)!. Maybe I don't understand what you mean, so maybe you could do an example with n=2 and m=3. – Arnaud Mortier Feb 07 '18 at 17:02
  • No combinatorial interpretation for the fraction (i.e., a set whose size is the fraction) is known (and people have been seeking for over 100 years). The closest things we have are a recursion with integer coefficients and an expression as a sum of integers. But both the recursion and the sum contain minus signs, so these don't translate into combinatorics. See Exercise 2.17 in Darij Grinberg, Notes on the combinatorial fundamentals of algebra, version of 22 January 2018, or google for "super-Catalan numbers". – darij grinberg Feb 07 '18 at 17:18

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