China Girls' Mathematical Olympiad

There are at most six permutations for any three numbers $x,y,z$. It suffices to show that there are only finitely many triples $(a,b,c)$, with $a\ge b\ge c$, of positive integers satisfying the equation $abc=2009(a+b+c)$. It follows that $abc\le 2009\times (3a)$ or $bc\le 2009\times 3=6027$. Clearly, there are finitely many pairs $(b,c)$ of positive integers satisfying the equation $bc\le 6027$ and for each fixed pair of integers $(b,c)$ there is at most one positive integer $a$ satisfying the equation $abc=2009(a+b+c)$ (because it is a linear equation in $a$).