PUMaC 2013 · 组合（B 组） · 第 1 题

PUMaC 2013 — Combinatorics (Division B) — Problem 1

[ 3 ] Including the original, how many ways are there to rearrange the letters in PRINCETON so that no two vowels (I, E, O) are consecutive and no three consonants (P, R, N, C, T, N) are consecutive? 2014

解析

[ 3 ] Including the original, how many ways are there to rearrange the letters in PRINCETON so that no two vowels (I, E, O) are consecutive and no three consonants (P, R, N, C, T, N) are consecutive? Solution We have six consonants and three vowels. The consonants then occur in at most four segments sectioned off by vowels. If only three segments have consonants in them, the only possibilities are CCVCCVCCV and VCCVCCVCC. If all four have consonants, we have four choose two, or six, ways to add the last two consonants. In all, we have 6 + 2 = 8 layouts, and 6! × 3! = 360(6) = 2160 ways to fill the layouts with letters. So the total is 2160(8) = 17280. 2 2014