Difference between revisions of "2013 AIME I Problems/Problem 4"
(→Solution) |
|||
Line 17: | Line 17: | ||
<math>\frac{3}{\binom{13}{5}}</math> = <math>\frac{1}{429}</math> , so ''<math>n</math>'' = <math>\boxed{429}</math> | <math>\frac{3}{\binom{13}{5}}</math> = <math>\frac{1}{429}</math> , so ''<math>n</math>'' = <math>\boxed{429}</math> | ||
+ | ==Video Solution== | ||
+ | |||
+ | https://www.youtube.com/watch?v=9way8JrtD04&t=555s | ||
+ | ~Shreyas S | ||
== See also == | == See also == | ||
{{AIME box|year=2013|n=I|num-b=3|num-a=5}} | {{AIME box|year=2013|n=I|num-b=3|num-a=5}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 11:30, 9 July 2020
Contents
[hide]Problem 4
In the array of 13 squares shown below, 8 squares are colored red, and the remaining 5 squares are colored blue. If one of all possible such colorings is chosen at random, the probability that the chosen colored array appears the same when rotated 90 degrees around the central square is , where n is a positive integer. Find n.
Solution
When the array appears the same after a 90-degree rotation, the top formation must look the same as the right formation, which looks the same as the bottom one, which looks the same as the right one. There are four of the same configuration. There are not enough red squares for these to be all red, nor are there enough blue squares for there to be more than one blue square in each three-square formation. Thus there are 2 reds and 1 blue in each, and a blue in the center. There are 3 ways to choose which of the squares in the formation will be blue, leaving the other two red.
There are ways to have 5 blue squares in an array of 13.
= , so =
Video Solution
https://www.youtube.com/watch?v=9way8JrtD04&t=555s ~Shreyas S
See also
2013 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
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.