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 | + | 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>. | ||
~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)== |
Latest revision as of 19:31, 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
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:
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:
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
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 |
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