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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>.
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==Video solution by [[User:grogg007|grogg007]]==
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https://youtu.be/wib5vos7Sd4?t=27
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==Solution 1==
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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, <math>AB</math> <math>CH</math> <math>DI</math> <math>EJ</math> <math>FK</math> <math>GL</math> should be counted equally as <math>HC</math> <math>AB</math> <math>DI</math> <math>EJ</math> <math>FK</math> <math>GL</math>.
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We construct two cases: <math>G</math> is the first letter of the last word and <math>G</math> is the second letter of the last word.
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Our first case is when <math>G</math> is the first letter of the last word. Then the second letter of the last word must be one of <math>H, I, J, K, L</math>. Call that set of <math>5</math> letters <math>\Omega</math>. There are <math>5</math> ways to choose the second letter from <math>\Omega</math>. The other <math>4</math> letters of <math>\Omega</math> must be used in the other <math>5</math> words.
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For the other 5 words, each of their first letters must be before <math>G</math> in the alphabet. Otherwise, the word with <math>G</math> will not be the last. There are <math>6</math> letters before <math>G</math>: <math>A,B,C,D,E,F</math>. Call that set of <math>6</math> letters <math>\Sigma</math>. Exactly one of the words must have two letters from <math>\Sigma</math>. The other 4 will have their first letter from <math>\Sigma</math> and the second letter from <math>\Omega</math>. There are <math>4!</math> ways to determine the possible pairings of letters from <math>\Sigma</math> and <math>\Omega</math>, respectively.
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Therefore, this case has <math>5 \cdot {6\choose{2}} \cdot 4! = 5 \cdot 15 \cdot 24 = 1800</math> orderings.
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The second case is when <math>G</math> is the second letter of the last word. You can see that the first letter of that word must be <math>F</math>. Otherwise, that word cannot be the last word. The other <math>5</math> words must start with <math>A</math>, <math>B</math>, <math>C</math>, <math>D</math>, and <math>E</math>. The second letter of each of those words will come from <math>\Omega</math>. There will be <math>5!</math> ways to distribute the elements of <math>\Omega</math> to one of <math>A, B, C, D, E</math>. There are therefore <math>5! = 120</math> orderings in the case.
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In total, there are <math>1800+120 = 1920</math> orderings. However, we want the probability. The number of ways to put the <math>12</math> letters into pairs is <math>11 \cdot 9 \cdot 7 \cdot 5 \cdot 3 \cdot 1</math>. This is true because we can say this: Start with <math>A</math>. It has <math>11</math> options for who it will partner with. There are now <math>10</math> letters left. Pick one of those letters. It has <math>9</math> options for who it will partner with. There are now <math>8</math> letters left. Continue until there are only <math>2</math> letters left, and there is only <math>1</math> option for that last word. Therefore, there will be  <math>11 \cdot 9 \cdot 7 \cdot 5 \cdot 3 \cdot 1</math> options.
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The probability is therefore <math>\frac{1920}{11 \cdot 9 \cdot 7 \cdot 5 \cdot 3 \cdot 1} = \frac{128}{693}</math>. The requested answer is <math>128 + 693 = \boxed{821}</math>.
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~lprado
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Minor latex edits by [[User:T3chn0b14d3|T3CHN0B14D3]]
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==Solution 2: Same but quicker==
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Splitting up into <math>2</math> cases: <math>G</math> is the first letter or the second letter of the last word.
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Case <math>1:</math> <math>G</math> in first letter
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Notice that <math>A</math> must take the first letter of first word, one of the letters <math>B</math> - <math>F</math> needs to be the second letter of a word and the rest being the first letter of a word.
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The combinations will be <math>1 + 2 + 3 + 4 + 5 = 15.</math> After the first <math>7</math> letters has been decided then the last <math>5</math> will just fill by <math>5!.</math> This case will have <math>15 \cdot 5!</math> outcomes.
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Case <math>2:</math> <math>G</math> in last letter
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Notice that <math>A</math> - <math>G</math> has been arranged by <math>A? B? C? D? E? FG,</math> where the <math>?</math> is undecided. We have another <math>5!</math> to fill out the possible outcomes.
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In total, there are <math>16 \cdot 5!.</math> The total case will be <math>11 \cdot 9 \cdot 7 \cdot 5 \cdot 3 \cdot 1</math> (Consider A must be in the first letter of first word, then you have <math>11</math> choices, then you must take the next letter in alphabetical order as mandatory, then you have a free choice of <math>9</math> and so on).
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Answer:
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<cmath>= \frac{16 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}{ 11 \cdot 9 \cdot 7 \cdot 5 \cdot 3 \cdot 1}</cmath>
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<cmath>= \frac{16 \cdot 4 \cdot 2}{11 \cdot 9 \cdot 7}</cmath>
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<cmath>= \frac{128}{ 693}</cmath>
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Therefore it gives us the answer of <math>{128 + 693 = \boxed{821}.}</math>
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~Mitsuihisashi14
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~Latex by [[User:Mathkiddus|mathkiddus]]
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==Video Solution 1 by SpreadTheMathLove==
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https://www.youtube.com/watch?v=59Q5bKVtTq4
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==See also==
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{{AIME box|year=2025|num-b=6|num-a=8|n=I}}
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{{MAA Notice}}

Latest revision as of 21:06, 12 August 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 $m+n$.

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

See also

2025 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 6 		Followed by
Problem 8
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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