1990 AIME Problems/Problem 8

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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

Solution 1

From left to right, suppose that the columns are labeled $L,M,$ and $R,$ respectively.

Consider the string $LLLMMRRR$. Since the set of all letter arrangements is bijective to the set of all shooting orders, the answer is the number of letter arrangements, which is $\frac{8!}{3! \cdot 2! \cdot 3!} = \boxed{560}.$

~Azjps (Solution)


Video Solution

https://www.youtube.com/watch?v=NGfMLCRUs3c&t=7s ~ MathEx

See also

1990 AIME (ProblemsAnswer KeyResources)
Preceded by
Problem 7
Followed by
Problem 9
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions

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