Difference between revisions of "1990 AIME Problems/Problem 8"
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== Solution == | == Solution == | ||
Suppose that the columns are labeled <math>A</math>, <math>B</math>, and <math>C</math>. Consider the string <math>AAABBBCC</math>. Since the arrangements of the strings is bijective to the order of shooting, the answer is the number of ways to arrange the letters which is <math>\frac{8!}{3! \cdot 3! \cdot 2!} = \boxed{560}</math>. | Suppose that the columns are labeled <math>A</math>, <math>B</math>, and <math>C</math>. Consider the string <math>AAABBBCC</math>. Since the arrangements of the strings is bijective to the order of shooting, the answer is the number of ways to arrange the letters which is <math>\frac{8!}{3! \cdot 3! \cdot 2!} = \boxed{560}</math>. | ||
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+ | ==Video Solution== | ||
+ | https://www.youtube.com/watch?v=NGfMLCRUs3c&t=7s ~ MathEx | ||
== See also == | == See also == |
Revision as of 20:06, 28 June 2020
Contents
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
Suppose that the columns are labeled , , and . Consider the string . Since the arrangements of the strings is bijective to the order of shooting, the answer is the number of ways to arrange the letters which is .
Video Solution
https://www.youtube.com/watch?v=NGfMLCRUs3c&t=7s ~ MathEx
See also
1990 AIME (Problems • Answer Key • Resources) | ||
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.