2025 AIME I Problems/Problem 7

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 $m+n$ .

Video solution by grogg007

https://youtu.be/wib5vos7Sd4?t=27

Solution 1

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

Minor latex edits by T3CHN0B14D3

Solution 2: Same but quicker

Splitting up into $2$ cases: $G$ is the first letter or the second letter of the last word.

Case $1:$ $G$ in first letter

Notice that $A$ must take the first letter of first word, one of the letters $B$ - $F$ needs to be the second letter of a word and the rest being the first letter of a word. The combinations will be $1 + 2 + 3 + 4 + 5 = 15.$ After the first $7$ letters has been decided then the last $5$ will just fill by $5!.$ This case will have $15 \cdot 5!$ outcomes.

Case $2:$ $G$ in last letter

Notice that $A$ - $G$ has been arranged by $A? B? C? D? E? FG,$ where the $?$ is undecided. We have another $5!$ to fill out the possible outcomes.

In total, there are $16 \cdot 5!.$ The total case will be $11 \cdot 9 \cdot 7 \cdot 5 \cdot 3 \cdot 1$ (Consider A must be in the first letter of first word, then you have $11$ choices, then you must take the next letter in alphabetical order as mandatory, then you have a free choice of $9$ and so on).

Answer: $= \frac{16 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}{ 11 \cdot 9 \cdot 7 \cdot 5 \cdot 3 \cdot 1}$ $= \frac{16 \cdot 4 \cdot 2}{11 \cdot 9 \cdot 7}$ $= \frac{128}{ 693}$ Therefore it gives us the answer of ${128 + 693 = \boxed{821}.}$

~Mitsuihisashi14 ~Latex by mathkiddus

Video Solution 1 by SpreadTheMathLove

https://www.youtube.com/watch?v=59Q5bKVtTq4

2025 AIME I (Problems • Answer Key • Resources)
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