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
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 ball of the same color, spots
,
,
,
,
, and
contain some permutation of all 6 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 balls; 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 blue balls. Similarly, there are
ways to place the blue balls, and so on, until there are
ways to place the purple balls. Thus, the probability is
, and the desired answer extraction is
.
~A1001
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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