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Difference between revisions of "2025 AIME I Problems/Problem 7"
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==Problem==
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The twelve letters <math>A</math>,<math>B</math>,<math>C</math>,<math>D</math>,<math>E</math>,<math>F</math>,<math>G</math>,<math>H</math>,<math>I</math>,<math>J</math>,<math>K</math>, and <math>L</math> are randomly grouped into six pairs of letters. The two letters in each pair are placed next to each other in alphabetical order to form six two-letter words, and then those six words are listed alphabetically. For example, a possible result is <math>AB</math>, <math>CJ</math>, <math>DG</math>, <math>EK</math>, <math>FL</math>, <math>HI</math>. The probability that the last word listed contains <math>G</math> is <math>\frac mn</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>.
 
==Solution==
 
==Solution==
 
Note that order does not matter here. This is because any permutation of the <math>6</math> pairs will automatically get ordered in alphabetical order. The same is true for within each of the pairs. In other words, AB CH DI EJ FK GL should be counted equally as HC AB DI EJ FK GL.
 
Note that order does not matter here. This is because any permutation of the <math>6</math> pairs will automatically get ordered in alphabetical order. The same is true for within each of the pairs. In other words, AB CH DI EJ FK GL should be counted equally as HC AB DI EJ FK GL.

Revision as of 19:19, 13 February 2025

Problem

The twelve letters $A$,$B$,$C$,$D$,$E$,$F$,$G$,$H$,$I$,$J$,$K$, and $L$ are randomly grouped into six pairs of letters. The two letters in each pair are placed next to each other in alphabetical order to form six two-letter words, and then those six words are listed alphabetically. For example, a possible result is $AB$, $CJ$, $DG$, $EK$, $FL$, $HI$. The probability that the last word listed contains $G$ is $\frac mn$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.

Solution

Note that order does not matter here. This is because any permutation of the $6$ pairs will automatically get ordered in alphabetical order. The same is true for within each of the pairs. In other words, AB CH DI EJ FK GL should be counted equally as HC AB DI EJ FK GL.

We construct two cases: $G$ is the first letter of the last word and $G$ is the second letter of the last word.

Our first case is when $G$ is the first letter of the last word. Then the second letter of the last word must be one of $H, I, J, K, L$. Call that set of $5$ letters $\Omega$. There are $5$ ways to choose the second letter from $\Omega$. The other $4$ letters of $\Omega$ must be used in the other $5$ words.

For the other 5 words, each of their first letters must be before $G$ in the alphabet. Otherwise, the word with $G$ will not be the last. There are $6$ letters before $G$: $A,B,C,D,E,F$. Call that set of $6$ letters $\Sigma$. Exactly one of the words must have two letters from $\Sigma$. The other 4 will have their first letter from $\Sigma$ and the second letter from $\Omega$. There are $4!$ ways to determine the possible pairings of letters from $\Sigma$ and $\Omega$, respectively.

Therefore, this case has $5 \cdot {6\choose{2}} \cdot 4! = 5 \cdot 15 \cdot 24 = 1800$ orderings.

The second case is when $G$ is the second letter of the last word. You can see that the first letter of that word must be $F$. Otherwise, that word cannot be the last word. The other $5$ words must start with $A$, $B$, $C$, $D$, and $E$. The second letter of each of those words will come from $\Omega$. There will be $5!$ ways to distribute the elements of $\Omega$ to one of $A, B, C, D, E$. There are therefore $5! = 120$ orderings in the case.

In total, there are $1800+120 = 1920$ orderings. However, we want the probability. The number of ways to put the $12$ letters into pairs is $11 \cdot 9 \cdot 7 \cdot 5 \cdot 3 \cdot 1$. This is true because we can say this: Start with $A$. It has $11$ options for who it will partner with. There are now $10$ letters left. Pick one of those letters. It has $9$ options for who it will partner with. There are now $8$ letters left. Continue until there are only $2$ letters left, and there is only $1$ option for that last word. Therefore, there will be $11 \cdot 9 \cdot 7 \cdot 5 \cdot 3 \cdot 1$ options.

The probability is therefore $\frac{1920}{11 \cdot 9 \cdot 7 \cdot 5 \cdot 3 \cdot 1} = \frac{128}{693}$. The requested answer is $128 + 693 = \boxed{821}$.

~lprado

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