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I have to prove that $$\frac{(2n)!(2m)!}{n!m!(n+m)!}$$ is always an integer.

I already have seen the same question here-Prove that for all non-negative integers $m,n$, $\frac{(2m)!(2n)!}{m!n!(m + n)!}$ is an integer.

But, I am again asking this question as I have a different method from that posted as answers and I am stuck mid way in my proof.So,I want some help to complete this proof. So, please don't close my question as a duplicate.

My attempt-

$$\frac{(2n)!(2m)!}{n!m!(n+m)!}$$

$$=\frac{n!(n+1)(n+2)...(n+n)(m!)(m+1)(m+2)...(m+m)}{n!m!(n+m)!}$$

$$=\frac{(n+1)(n+2)...(n+n)(m+1)(m+2)...(m+m)}{(n+m)!}$$

Case 1:-Let $n>m$

$$=\frac{(n+1)(n+2)...(n+m)(n+m+1)...(n+n)(m+1)(m+2)...(m+m)}{n!(n+1)(n+2)...(n+m)}$$

$$=\frac{(n+m+1)(n+m+2)...(n+n)(m+1)(m+2)...(m+m)}{n!}$$

Now,from $(n+m+1)$ to $(n+n)$, we have $n-m$ terms and from $(m+1)$ to $(m+m)$ we have $m$ terms.So,total we have $(n-m)+m=n$ terms.

Apparently, it seemed to me that this will be an integer as product of any $n$ consecutive integers is always divisible by $n!$.But then to may horror I found out that the $n$ terms in the numerator may not be consecutive.

So, how to get a way out of this by using be method (please, by using my approach, not by any other way!!)?

Thanks a lot in advance!!

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    $\begingroup$ I don't think there is an easy way to patch your non-proof. Why do you need a third alternative, since two working approaches are given in the answers to the linked question? $\endgroup$ – Jack D'Aurizio Jul 1 '16 at 15:34
  • $\begingroup$ @JackD'Aurizio We can have hundreds of ways for solving a problem....I was just thinking how it could done with my approach.... $\endgroup$ – tatan Jul 1 '16 at 15:36
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    $\begingroup$ @tatan: just because you've started down a path doesn't mean that path leads anywhere. $\endgroup$ – Cheerful Parsnip Jul 1 '16 at 15:44
  • $\begingroup$ @GrumpyParsnip Are you sure it does't lead anywhere? $\endgroup$ – tatan Jul 1 '16 at 15:45
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    $\begingroup$ @tatan I never claimed that the path led nowhere. I'm just pointing out that your point of view is rather at odds with the way mathematics works. You can't know a technique will work in advance, as your question seems to presuppose. $\endgroup$ – Cheerful Parsnip Jul 1 '16 at 17:22
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It is enough to show that $\binom{n+m}{n}=\binom{n+m}{m}$ is a divisor of $\binom{2n}{n}\cdot\binom{2m}{m}$.

Assume that we have a parliament with $2n$ politicians in the left wing and $2m$ politicians in the right wing. $\binom{2n}{n}\binom{2m}{m}$ is the number of committees made by exactly $n$ politicians from the left wing and exactly $m$ politicians from the right wing. Every subset of $n$ politicians in the left wing has the same probability to be part of the super-committee, and $$ \binom{n+m}{n}\mid \binom{2n}{n}\binom{2m}{m} $$ also comes from computing the probability for a single politician to be part of the super-committee.

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    $\begingroup$ How does this divisibility come from computing the probability? $\endgroup$ – darij grinberg Jul 1 '16 at 16:33

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