2022 AIME I Problems/Problem 9

Problem

Ellina has twelve blocks, two each of red $\left({\bf R}\right),$ blue $\left({\bf B}\right),$ yellow $\left({\bf Y}\right),$ green $\left({\bf G}\right),$ orange $\left({\bf O}\right),$ and purple $\left({\bf P}\right).$ 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 ${\text {\bf R B B Y G G Y R O P P O}}$ is even. Ellina arranges her blocks in a row in random order. The probability that her arrangement is even is $\frac mn$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ .

Solution 1

Consider this position chart: ${\text {\bf 1 2 3 4 5 6 7 8 9 10 11 12}}$ Since there has to be an even number of spaces between each ball 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: $\frac{6!\cdot6!}{\frac{12!}{(2!)^6}} = \frac{6!\cdot2^6}{7\cdot8\cdot9\cdot10\cdot11\cdot12} = \frac{2^6}{7\cdot11\cdot12} = \frac{16}{231}$ , which is in simplest form. So $m + n = 16 + 231 = \boxed{247}$ .

2022 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 8	Followed by Problem 10
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2022 AIME I Problems/Problem 9

Problem

Solution 1

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