Difference between revisions of "2022 AIME I Problems/Problem 9"
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==Solution 1== | ==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 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 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 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>. | which is in simplest form. So, <math>m + n = 16 + 231 = \boxed{247}</math>. | ||
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==Solution 2== | ==Solution 2== | ||
| − | We can simply use constructive counting. First, let us place the red | + | 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 | ~A1001 | ||
Latest revision as of 18:31, 31 January 2024
Contents
Problem
Ellina has twelve blocks, two each of red (
), blue (
), yellow (
), green (
), orange (
), and purple (
). Call an arrangement of blocks 
if there is an even number of blocks between each pair of blocks of the same color. For example, the arrangement
![\[\textbf{R B B Y G G Y R O P P O}\]](http://latex.artofproblemsolving.com/2/2/4/2246d3f2384b544ed1634ec69c8d0c4be9f303e6.png)
is even. Ellina arranges her blocks in a row in random order. The probability that her arrangement is even is 
where 
and 
are relatively prime positive integers. Find 
Solution 1
Consider this position chart: ![\[\textbf{1 2 3 4 5 6 7 8 9 10 11 12}\]](http://latex.artofproblemsolving.com/d/2/6/d268e4b5ac272cbb26d410153e143b268e09cfa4.png)
Since there has to be an even number of spaces between each pair of the same color, spots 
, 
, 
, 
, 
, and 
contain some permutation of all 
colored balls. Likewise, so do the even spots, so the number of even configurations is 
(after putting every pair of colored balls in opposite parity positions, the configuration can be shown to be even). This is out of 
possible arrangements, so the probability is: ![\[\frac{6!\cdot6!}{\frac{12!}{(2!)^6}} = \frac{6!\cdot2^6}{7\cdot8\cdot9\cdot10\cdot11\cdot12} = \frac{2^4}{7\cdot11\cdot3} = \frac{16}{231},\]](http://latex.artofproblemsolving.com/a/9/9/a995946c0ecb2b47d5a183478a43f48d08e23fe6.png)
which is in simplest form. So, 
.
~Oxymoronic15
Solution 2
We can simply use constructive counting. First, let us place the red blocks; choose the first slot in 
ways, and the second in ways, because the number is cut in half due to the condition in the problem. This gives 
ways to place the red blocks. Similarly, there are 
ways to place the blue blocks, and so on, until there are 
ways to place the purple blocks. Thus, the probability is ![\[\frac{12 \cdot 6 \cdot 10 \cdot 5 \cdot 8 \cdot 4 \cdot 6 \cdot 3 \cdot 4 \cdot 2 \cdot 2 \cdot 1}{12!}=\frac{16}{231},\]](http://latex.artofproblemsolving.com/b/8/6/b8612c1f874b70bedcda99bfd7b81767c662389a.png)
and the desired answer is 
.
~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 
. Then place any other block similarly, with 
ways (basic counting). You get then 
ways to place the blocks evenly, and 
ways to place the blocks in any way, so you get 
by simplifying.
-drag00n
Video Solution (Mathematical Dexterity)
https://www.youtube.com/watch?v=dkoF7StwtrM
Video Solution (Power of Logic)
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. 
