to simplify the problem, let's say the four evelopes are 1, 2, 3, and 4 and the corresponding envelopes are 1', 2', 3' and 4' (let's say the envolpes are put on table in this order, so we only need to consider the letters' location)
total ways of allocation: P(4,4)
exactly one letter in the right envelop: 4 choices,1-1', or 2-2'...
(let's say, we put letter 4 right, ie. 4-4')
1, 2 and 3 are left and we need to put them all wrong. We only have two choices, 2,3,1 or 3,1,2. So taotal ways of allocation with only one letter int he right envelope: 4*2.
So the possibility should be: (4*2)/p(4,4)=8/24=1/3.
A professional way to solve this problem is:
[p(1,4)*p(1,2)*1]/p(4,4). The above example can help you understand why it's
p(1,2)*1 after p(1,4)作者: steven.wei 时间: 2003-1-7 21:34
感谢二位仁兄的指点。一个在11点半，一个在凌晨。看来数学不拿5＊对不起帮助我的人，再次表示感谢。
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