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Difference between revisions of "2022 AIME I Problems/Problem 9"
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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>
 
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 ==
+
==Solution 1==
 
Consider this position chart: <cmath>\textbf{1 2 3 4 5 6 7 8 9 10 11 12}</cmath>
 
Consider this position chart: <cmath>\textbf{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{2^4}{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>
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~Oxymoronic15
 
~Oxymoronic15
 +
 +
==Solution 2==
 +
We can simply use constructive counting. First, let us place the red balls; choose the first slot in <math>12</math> ways, and the second in <math>6</math> ways, because the number is cut in half due to the condition in the problem. This gives <math>12 \cdot 6</math> ways to place the blue balls. Similarly, there are <math>10 \cdot 5</math> ways to place the blue balls, and so on, until there are <math>2 \cdot 1</math> ways to place the purple balls. Thus, the probability is <math>\frac{12 \cdot 6 \cdot 10 \cdot 5 \cdot 8 \cdot 4 \cdot 6 \cdot 3 \cdot 4 \cdot 2 \cdot 2 \cdot 1}{12!}=\frac{16}{123}</math>, and the desired answer extraction is <math>16+123=\boxed{\textbf{139}}</math>.
 +
 +
~A1001
  
 
==Video Solution (Mathematical Dexterity)==
 
==Video Solution (Mathematical Dexterity)==

Revision as of 13:52, 7 April 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 1

Consider this position chart: \[\textbf{1 2 3 4 5 6 7 8 9 10 11 12}\] 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

Solution 2

We can simply use constructive counting. First, let us place the red balls; choose the first slot in $12$ ways, and the second in $6$ ways, because the number is cut in half due to the condition in the problem. This gives $12 \cdot 6$ ways to place the blue balls. Similarly, there are $10 \cdot 5$ ways to place the blue balls, and so on, until there are $2 \cdot 1$ ways to place the purple balls. Thus, the probability is $\frac{12 \cdot 6 \cdot 10 \cdot 5 \cdot 8 \cdot 4 \cdot 6 \cdot 3 \cdot 4 \cdot 2 \cdot 2 \cdot 1}{12!}=\frac{16}{123}$, and the desired answer extraction is $16+123=\boxed{\textbf{139}}$.

~A1001

Video Solution (Mathematical Dexterity)

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

Video Solution (Power of Logic)

https://youtu.be/AF6TOG7MSwA

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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