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

Revision as of 16:01, 13 January 2023

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 pair 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 red 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 $\frac{12 \cdot 6 \cdot 10 \cdot 5 \cdot 8 \cdot 4 \cdot 6 \cdot 3 \cdot 4 \cdot 2 \cdot 2 \cdot 1}{12!}=\frac{16}{231},$ and the desired answer extraction is $16+231=\boxed{247}$ .

~A1001

Solution 3

Use constructive counting, as per above. WLOG, place the red blocks first. There are 11 ways to place them with distance 0, 9 ways them to place with distance 2, so on, so the way to place red blocks is $11+9+7+5+3+1=36$ . Then place any other block similarly, with $25$ ways (basic counting). You get then $6!^2$ ways to place the blocks evenly, and $12!/64$ ways to place the blocks in any way, so you get $\frac{16}{231}=247$ by simplifying.

-drag00n

Video Solution (Mathematical Dexterity)

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

Video Solution (Power of Logic)

https://youtu.be/AF6TOG7MSwA

@@ Line 1: / Line 1: @@
-.
+== Problem ==
+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>\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==
+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 pair 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 <math>6</math> 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>.
+~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 red 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 <cmath>\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}{231},</cmath> and the desired answer extraction is <math>16+231=\boxed{247}</math>.
+~A1001
+==Solution 3==
+Use constructive counting, as per above. WLOG, place the red blocks first. There are 11 ways to place them with distance 0, 9 ways them to place with distance 2, so on, so the way to place red blocks is <math>11+9+7+5+3+1=36</math>. Then place any other block similarly, with <math>25</math> ways (basic counting). You get then <math>6!^2</math> ways to place the blocks evenly, and <math>12!/64</math> ways to place the blocks in any way, so you get <math>\frac{16}{231}=247</math> by simplifying.
+-drag00n
+==Video Solution (Mathematical Dexterity)==
+https://www.youtube.com/watch?v=dkoF7StwtrM
+==Video Solution (Power of Logic)==
+https://youtu.be/AF6TOG7MSwA
+==See Also==
+{{AIME box|year=2022|n=I|num-b=8|num-a=10}}
+[[Category:Intermediate Combinatorics Problems]]
+{{MAA Notice}}

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