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标题: 甲乙丙三人传球 [打印本页]

作者: 野菜花 时间: 2005-4-28 04:09
标题: 甲乙丙三人传球

由甲开始， 5次传球后球仍回到甲手中的不同方式共有多少种？

作者: fzy 时间: 2005-4-28 18:23
标题: 回复：甲乙丙三人传球

To have the ball back to 甲 after n passes, the ball must be at some one else's hand after n-1 passes. Let P(n) be the number, then we have in general P(1) = 0 and P(n) = 2^(n-1) - P(n-1). SO it is easy to calculate that P(5) = 10. The general format is P(n) = (2^n + 2(-1)^n) / 3. (Never can get rid of induction! )

作者: 野菜花 时间: 2005-4-28 21:22
标题: Great! [@};-][@};-]

You dislike induction, but you are very good at it!

You have generalize 5 passes to n and got an excellent formula.

作者: 野菜花 时间: 2005-4-29 00:01
标题: 回复：回复：甲乙丙三人传球

So, we can get a recursive formula for k people and n passes:

P(n)=(k-1)^(n-1)-P(n-1), right ?

作者: fzy 时间: 2005-4-29 00:10
标题: 回复：回复：回复：甲乙丙三人传球

Right. And general formula: P(n)=((k-1)^n + (k-1)*(-1)^n) / k.

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