Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Video Solution (Mathematical Dexterity))
m (→‎Solution 2)
 
(15 intermediate revisions by 7 users not shown)
Line 1: Line 1:
 
== Problem ==
 
== Problem ==
−
Ellina has twelve blocks, two each of red <math>\left({\bf R}\right),</math> blue <math>\left({\bf B}\right),</math> yellow <math>\left({\bf Y}\right),</math> green <math>\left({\bf G}\right),</math> orange <math>\left({\bf O}\right),</math> and purple <math>\left({\bf P}\right).</math> 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
+
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> {\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>\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 ==
+
==Solution 1==
−
Consider this position chart:
+
Consider this position chart: <cmath>\textbf{1 2 3 4 5 6 7 8 9 10 11 12}</cmath>
−
<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 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>
−
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>.
−
, which is in simplest form. So <math>m + n = 16 + 231 = \boxed{247}</math>.
 
  
−
-Oxymoronic15
+
~Oxymoronic15
 +
 
 +
==Solution 2==
 +
We can simply use constructive counting. First, let us place the red blocks; 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 blocks. Similarly, there are <math>10 \cdot 5</math> ways to place the blue blocks, and so on, until there are <math>2 \cdot 1</math> ways to place the purple blocks. 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 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)==
 
==Video Solution (Mathematical Dexterity)==
Line 19: Line 30:
 
==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}}
 +
 +
[[Category:Intermediate Combinatorics Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 18:31, 31 January 2024

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 blocks; 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 blocks. Similarly, there are $10 \cdot 5$ ways to place the blue blocks, and so on, until there are $2 \cdot 1$ ways to place the purple blocks. 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 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

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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2022_AIME_I_Problems/Problem_9&oldid=213986"