Problem:
The twelve letters and 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 . The probability that the last word listed contains is , where and are relatively prime positive integers. Find .
Solution:
Case on whether is in the first or second slot of its pair.
If is in the first slot of its pair, one of the letters must follow it. Of the remaining letters, they must not pair to each other and must pair to letters in the set only. There are choices for the letter to follow and choices to pick the letters pairing to the remaining , giving a total of ways for this case.
If is in the second slot of its pair, must be the letter preceding it. If or any earlier letter precedes it, it will never be the last word in alphabetical order. The letters must pair to the set in some permutation to ensure that is the last word, giving ways for this case.
For the denominator, there are ways to choose the partner of , ways to choose the partner of the earliest remaining letter, ways to choose the partner of the next earliest remaining letter ... etc. for a total of ways in total. The final probability is
The problems on this page are the property of the MAA's American Mathematics Competitions