PUMaC 2016 · 组合（A 组） · 第 2 题

PUMaC 2016 — Combinatorics (Division A) — Problem 2

32 teams, ranked 1 through 32, enter a basketball tournament that works as follows: the teams are randomly paired and in each pair, the team that loses is out of the competition. The remaining 16 teams are randomly paired, and so on, until there is a winner. A higher- ranked team always wins against a lower-ranked team. If the probability that the team ranked 3 (the third-best team) is one of the last four teams remaining can be written in simplest form m as , compute m + n . n

解析

This is the same as putting the teams in a bracket-style tournament at random. The probability that the teams ranked 1 and 2 are not in the same quarter of the draw as the team ranked 3 24 · 23 92 is the relevant probability, and it is = , so the answer is 92 + 155 = 247 . 31 · 30 155 Problem written by Eric Neyman.