Difference between revisions of "2022 AIME I Problems/Problem 9"
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== Solution == | == Solution == | ||
| − | Consider this position chart: | + | Consider this position chart: <cmath>\textbf{1 2 3 4 5 6 7 8 9 10 11 12}</cmath> |
| − | <cmath> | ||
Since there has to be an even number of spaces between each ball of the same color, spots <math>1</math>, <math>3</math>, <math>5</math>, <math>7</math>, <math>9</math>, and <math>11</math> contain some permutation of all 6 colored balls. Likewise, so do the even spots, so the number of even configurations is <math>6! \cdot 6!</math> (after putting every pair of colored balls in opposite parity positions, the configuration can be shown to be even). This is out of <math>\frac{12!}{(2!)^6}</math> possible arrangements, so the probability is: <cmath>\frac{6!\cdot6!}{\frac{12!}{(2!)^6}} = \frac{6!\cdot2^6}{7\cdot8\cdot9\cdot10\cdot11\cdot12} = \frac{2^4}{7\cdot11\cdot3} = \frac{16}{231},</cmath> | Since there has to be an even number of spaces between each ball of the same color, spots <math>1</math>, <math>3</math>, <math>5</math>, <math>7</math>, <math>9</math>, and <math>11</math> contain some permutation of all 6 colored balls. Likewise, so do the even spots, so the number of even configurations is <math>6! \cdot 6!</math> (after putting every pair of colored balls in opposite parity positions, the configuration can be shown to be even). This is out of <math>\frac{12!}{(2!)^6}</math> possible arrangements, so the probability is: <cmath>\frac{6!\cdot6!}{\frac{12!}{(2!)^6}} = \frac{6!\cdot2^6}{7\cdot8\cdot9\cdot10\cdot11\cdot12} = \frac{2^4}{7\cdot11\cdot3} = \frac{16}{231},</cmath> | ||
which is in simplest form. So, <math>m + n = 16 + 231 = \boxed{247}</math>. | which is in simplest form. So, <math>m + n = 16 + 231 = \boxed{247}</math>. | ||
Revision as of 18:46, 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
![\[\textbf{R B B Y G G Y R O P P O}\]](http://latex.artofproblemsolving.com/2/2/4/2246d3f2384b544ed1634ec69c8d0c4be9f303e6.png)
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
Consider this position chart: ![\[\textbf{1 2 3 4 5 6 7 8 9 10 11 12}\]](http://latex.artofproblemsolving.com/d/2/6/d268e4b5ac272cbb26d410153e143b268e09cfa4.png)
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: ![\[\frac{6!\cdot6!}{\frac{12!}{(2!)^6}} = \frac{6!\cdot2^6}{7\cdot8\cdot9\cdot10\cdot11\cdot12} = \frac{2^4}{7\cdot11\cdot3} = \frac{16}{231},\]](http://latex.artofproblemsolving.com/a/9/9/a995946c0ecb2b47d5a183478a43f48d08e23fe6.png)
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. 
