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Combinatorics Difficulty 5.8 AIME, harder Find the answer United States

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 $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ .

Your answer

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

There are $11 \cdot 9 \cdot 7 \cdot 5 \cdot 3 \cdot 1$ equally likely ways for the letters to be paired. This can be seen by considering the successive choices of a partner for the unpaired letter that comes first alphabetically. The letter $G$ is the first letter in its pair, and its pair is listed last if $G$ is paired with $H$ , $I$ , $J$ , $K$ , or $L$ (5 choices), two of $A$ , $B$ , $C$ , $D$ , $E$ , and $F$ are paired with each other ( $\binom{6}{2} = 15$ choices), and the 4 remaining early letters are paired with the 4 remaining late letters ( $4! = 24$ choices). The letter $G$ is the second letter in its pair, and its pair is listed last if $G$ is paired with $F$ and each of $A$ through $E$ is paired with one of $H$ through $L$ . This can happen in $5! = 120$ ways. Therefore
$\frac{m}{n} = \frac{5 \cdot 15 \cdot 24 + 120}{11 \cdot 9 \cdot 7 \cdot 5 \cdot 3 \cdot 1} = \frac{15 \cdot 8 + 8}{11 \cdot 9 \cdot 7} = \frac{128}{693}.$
The requested sum is $128 + 693 = 821$ .

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.