Difference between revisions of "2022 AIME I Problems/Problem 9"

Revision as of 18:45, 23 February 2022

Problem

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

Solution

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: $\frac{6!\cdot6!}{\frac{12!}{(2!)^6}} = \frac{6!\cdot2^6}{7\cdot8\cdot9\cdot10\cdot11\cdot12} = \frac{2^4}{7\cdot11\cdot3} = \frac{16}{231},$ which is in simplest form. So, $m + n = 16 + 231 = \boxed{247}$ .

-Oxymoronic15

Video Solution (Mathematical Dexterity)

https://www.youtube.com/watch?v=dkoF7StwtrM

Video Solution (Power of Logic)

https://youtu.be/AF6TOG7MSwA

@@ Line 4: / Line 4: @@
 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 ==
 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>
+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>.
 -Oxymoronic15

2022 AIME I (Problems • Answer Key • Resources)
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