Difference between revisions of "2022 AIME I Problems/Problem 9"
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== Problem == | == Problem == | ||
− | Ellina has twelve blocks, two each of red <math>\ | + | Ellina has twelve blocks, two each of red (<math>\textbf{R}</math>), blue (<math>\textbf{B}</math>), yellow (<math>\textbf{Y}</math>), green (<math>\textbf{G}</math>), orange (<math>\textbf{O}</math>), and purple (<math>\textbf{P}</math>). Call an arrangement of blocks <math>\textit{even}</math> if there is an even number of blocks between each pair of blocks of the same color. For example, the arrangement |
− | <cmath> | + | <cmath>\textbf{R B B Y G G Y R O P P O}</cmath> |
+ | is even. Ellina arranges her blocks in a row in random order. The probability that her arrangement is even is <math>\frac{m}{n},</math> where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n.</math> | ||
== Solution 1 == | == Solution 1 == |
Revision as of 18:43, 23 February 2022
Contents
Problem
Ellina has twelve blocks, two each of red (), blue (), yellow (), green (), orange (), and purple (). Call an arrangement of blocks if there is an even number of blocks between each pair of blocks of the same color. For example, the arrangement is even. Ellina arranges her blocks in a row in random order. The probability that her arrangement is even is where and are relatively prime positive integers. Find
Solution 1
Consider this position chart: Since there has to be an even number of spaces between each ball of the same color, spots , , , , , and contain some permutation of all 6 colored balls. Likewise, so do the even spots, so the number of even configurations is (after putting every pair of colored balls in opposite parity positions, the configuration can be shown to be even). This is out of possible arrangements, so the probability is: , which is in simplest form. So .
-Oxymoronic15
Video Solution (Mathematical Dexterity)
https://www.youtube.com/watch?v=dkoF7StwtrM
Video Solution (Power of Logic)
See Also
2022 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 8 |
Followed by Problem 10 | |
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.