HMMT 二月 2005 · 冲刺赛 · 第 4 题
HMMT February 2005 — Guts Round — Problem 4
题目详情
英文原题
- [6] What is the probability that in a randomly chosen arrangement of the numbers andletters in “HMMT2005,” one can read either “HMMT” or “2005” from left to right?
(For example, in “5HM0M20T,” one can read “HMMT.”)
解析
英文解析
- What is the probability that in a randomly chosen arrangement of the numbers andletters in “HMMT2005,” one can read either “HMMT” or “2005” from left to right?
(For example, in “5HM0M20T,” one can read “HMMT.”)
Solution: 23 / 144
( )1
4!8
To read “HMMT,” there are ways to place the letters, and ways to place the
4 2
( )
4!8
numbers. Similarly, there are arrangements where one can read ”2005.” The
4 2
( )
number of arrangements in which one can read both is just . The total number of 8
8!4
arrangements is , thus the answer is
( ) ( ) ( )4
( )
8 8 8
4! 4!
- −
8 4 23
4 2 4 2 4 = · 23 = .
8!
4 8! 144 4