HMMT 二月 2009 · COMB 赛 · 第 3 题
HMMT February 2009 — COMB Round — Problem 3
题目详情
英文原题
- [ 4 ] How many rearrangements of the letters of “HMMTHMMT” do not contain the substring “HMMT”?
(For instance, one such arrangement is HMMHMTMT.)
解析
英文解析
- [ 4 ] In how many ways can you rearrange the letters of “HMMTHMMT” such that the consecutivesubstring “HMMT” does not appear?
Answer: 361
Solution: There are 8! / (4!2!2!) = 420 ways to order the letters. If the permuted letters contain
“HMMT”, there are 5 · 4! / 2! = 60 ways to order the other letters, so we subtract these. However, wehave subtracted “HMMTHMMT” twice, so we add it back once to obtain 361 possibilities.