In a shooting match, eight clay targets are arranged in two hanging columns of three targets each and one column of two targets. A marksman is to break all the targets according to the following rules: 
1) The marksman first chooses a column from which a target is to be broken. 
2) The marksman must then break the lowest remaining target in the chosen column. 
If the rules are followed, in how many different orders can the eight targets be broken?

[1990 AIME Problem 8.png](https://artofproblemsolving.com/wiki/index.php/File:1990_AIME_Problem_8.png)
Clearly, the marksman must shoot the left column three times, the middle column two times, and the right column three times.
From left to right, suppose that the columns are labeled $L,M,$ and $R,$ respectively. We consider the string $LLLMMRRR:$
Since the letter arrangements of $LLLMMRRR$ and the shooting orders have one-to-one correspondence, we count the letter arrangements: \[\frac{8!}{3!\cdot2!\cdot3!} = \boxed{560}.\]
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