Difference between revisions of "1990 AIME Problems/Problem 8"

Revision as of 10:41, 19 June 2021

Problem

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?

Solution

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 $\frac{8!}{3!\cdot 2!\cdot3!} = \boxed{560}.$

~Azjps (Solution)

~MRENTHUSIASM (Revision)

Remark

We can count the letter arrangements of $LLLMMRRR$ in two ways:

~MRENTHUSIASM

Video Solution

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

@@ Line 13: / Line 13: @@
 From left to right, suppose that the columns are labeled <math>L,M,</math> and <math>R,</math> respectively.
-Consider the string <math>LLLMMRRR.</math> 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 <math>\frac{8!}{3! \cdot 2! \cdot 3!} = \boxed{560}.</math>
+Consider the string <math>LLLMMRRR.</math> 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 <math>\frac{8!}{3!\cdot 2!\cdot3!} = \boxed{560}.</math>
 ~Azjps (Solution)

1990 AIME (Problems • Answer Key • Resources)
Preceded by Problem 7	Followed by Problem 9
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