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HMMT 二月 2004 · COMB 赛 · 第 6 题

HMMT February 2004 — COMB Round — Problem 6

专题
Contest Math / 竞赛数学
难度
L3
来源
HMMT

题目详情

英文原题

  1. A committee of 5 is to be chosen from a group of 9 people. How many ways can it bechosen, if Bill and Karl must serve together or not at all, and Alice and Jane refuse toserve with each other?
解析

英文解析

  1. A committee of 5 is to be chosen from a group of 9 people. How many ways can it bechosen, if Bill and Karl must serve together or not at all, and Alice and Jane refuse toserve with each other?
    Solution: 41
    ( )
    If Bill and Karl are on the committee, there are = 35 ways for the other group 7
    members to be chosen. However, if Alice and Jane are on the committee with Bill and 3
    ( )
    Karl, there are = 5 ways for the last member to be chosen, yielding 5 unacceptable 5
    ( )1
    committees. If Bill and Karl are not on the committee, there are = 21 ways for 7
    the 5 members to be chosen, but again if Alice and Jane were to be on the committee,5
    ( )
    there would be = 10 ways to choose the other three members, yielding 10 more 5
    unacceptable committees. So, we obtain (35 − 5) + (21 − 10) = 41 ways the committee 3
    can be chosen.