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Difference between revisions of "2022 AIME I Problems/Problem 9"
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<cmath> {\text {\bf 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 mn</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>.
 
<cmath> {\text {\bf 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 mn</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>.
  
 +
== Solution 1 ==
 +
Consider this position chart:
 +
<cmath> {\text {\bf 1 2 3 4 5 6 7 8 9 10 11 12}}</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{2^6}{7\cdot11\cdot12} = \frac{16}{231}</cmath>
 +
, which is in simplest form. So <math>m + n = 16 + 231 = \boxed{247}</math>.
 
==See Also==
 
==See Also==
 
{{AIME box|year=2022|n=I|num-b=8|num-a=10}}
 
{{AIME box|year=2022|n=I|num-b=8|num-a=10}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 21:11, 17 February 2022

Problem

Ellina has twelve blocks, two each of red $\left({\bf R}\right),$ blue $\left({\bf B}\right),$ yellow $\left({\bf Y}\right),$ green $\left({\bf G}\right),$ orange $\left({\bf O}\right),$ and purple $\left({\bf P}\right).$ Call an arrangement of blocks even if there is an even number of blocks between each pair of blocks of the same color. For example, the arrangement \[{\text {\bf R B B Y G G Y R O P P O}}\]is even. Ellina arranges her blocks in a row in random order. The probability that her arrangement is even is $\frac mn$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.

Solution 1

Consider this position chart: \[{\text {\bf 1 2 3 4 5 6 7 8 9 10 11 12}}\] Since there has to be an even number of spaces between each ball of the same color, spots $1$, $3$, $5$, $7$, $9$, and $11$ contain some permutation of all 6 colored balls. Likewise, so do the even spots, so the number of even configurations is $6! \cdot 6!$ (after putting every pair of colored balls in opposite parity positions, the configuration can be shown to be even). This is out of $\frac{12!}{(2!)^6}$ possible arrangements, so the probability is: \[\frac{6!\cdot6!}{\frac{12!}{(2!)^6}} = \frac{6!\cdot2^6}{7\cdot8\cdot9\cdot10\cdot11\cdot12} = \frac{2^6}{7\cdot11\cdot12} = \frac{16}{231}\] , which is in simplest form. So $m + n = 16 + 231 = \boxed{247}$.

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

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