Prove that if $a$ and $b$ are natural numbers, then $a!b!|(a+b)!$

Asked Apr 17 '19 at 11:41

Active Apr 17 '19 at 11:51

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I was reading the first chapter (about divisibility) of Elementary Theory of Numbers by W.Sierpinski and I stumbled upon the following exercise:

Prove that if $a$ and $b$ are natural numbers, then $a!b!|(a+b)!$

the book provide an answer that rely on the fact that the theorem is true if at least one of the numbers $a$ and $b$ is equal to 1, since for each natural $b$ we have $(b+1)!=b!(b+1)$, whence $1!b!|(1+b)!$. And by using induction the theorem is proved to be true for all natural $a$ and $b$.

I'm curious if there is annother answer that doesn't rely on induction.

asked Apr 17 '19 at 11:41

sifadil

Possibly of interest: Does $a!b!$ always divide $(a+b)!$ – Minus One-Twelfth Apr 17 '19 at 11:44
2

If you can use that binomial coefficient ${n\choose m}$ is integer for all $m,n\geq 0,m\leq n$ then it follows almost immediately from that fact. – kingW3 Apr 17 '19 at 11:46

Prove that if $a$ and $b$ are natural numbers, then $a!b!|(a+b)!$

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