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标题: 求解一道数学题 [打印本页]

作者: teddyabc 时间: 2011-2-9 21:31 标题: 求解一道数学题

If a committee of 3 people is to be selected from among 5 married couples so that the committee does not include two people who are married to each other, how many such committees are possible?

A.20
B.40
C.50
D.80
E.120
答案是D
而我得出的结果却是10*8*6=480
相差甚远啊，NN们能否一解释疑啊

作者: avalon9527 时间: 2011-2-10 07:05

这类题目不能像你那么解答，你那么解答是有很多重复的。
这类题应该先选哪一对，再选这一对的哪一个人，实际上应该是C(3,5)*C(2,1)*C(2,1)*C(2,1)=80 这样才不会有重复。

作者: 卮言浅夏 时间: 2011-2-10 14:37

480/3! = 80

The three people you selected as ABC would be the same composition as BCA, CAB, etc.

作者: xiexiang 时间: 2011-2-11 09:04

5对选3对， 10
然后三对里各选1个 2*2*2
10*8=80

你这种有大量重复
如1a2a3a 和2a3a1a 岂不是一样吗

作者: teddyabc 时间: 2011-2-12 18:41

原来是这个意思，灰常感谢！！

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