by aam » Tue Apr 23, 2013 5:06 pm
We can say:
(m+n)!=(m+n)(m+n-1)(m+n-2)...(m+n-(n-1)) x m!
clearly (m+n)! is dividable by m!, and we got:
(m+n)!/m!=(m+n)(m+n-1)(m+n-2)...(m+n-(n-1)).
Now, the right hand side of the last equation consists of multiplication of n consecutive integers. Among them, there exist integers dividable by n, n-1, n-2, ... , 1. So, the right hand side is dividable by n(n-1)(n-2)...1 = n!,
And we can conclude (m+n)! is dividable by m!n!.