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 $a$).