Difference between revisions of "1972 IMO Problems/Problem 3"

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Let m and n be arbitrary non-negative integers. Prove that
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Let <math>m</math> and <math>n</math> be arbitrary non-negative integers. Prove that
 
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<cmath>\frac{(2m)!(2n)!}{m!n!(m+n)!}</cmath>
<math>((2m)!(2n)!)/mn!(m+n)!</math>
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is an integer. (<math>0! = 1</math>.)
 
 
is an integer. (0! = 1.)
 
  
 
== Solution ==
 
== Solution ==

Revision as of 15:38, 17 October 2014

Let $m$ and $n$ be arbitrary non-negative integers. Prove that \[\frac{(2m)!(2n)!}{m!n!(m+n)!}\] is an integer. ($0! = 1$.)

Solution