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[组合] 通过计算概率得出整除性

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hbghlyj Post time 2023-10-13 22:25 |Read mode
本帖最后由 hbghlyj 于 2023-11-8 11:34 编辑 IMO 1972, Problem 3.
关于@Jack D'Aurizio的回答
Every subset of $n$ politicians in the left wing has the same probability to be part of the super-committee, and
$$ \binom{n+m}{n}\mid \binom{2n}{n}\binom{2m}{m} $$
also comes from computing the probability for a single politician to be part of the super-committee.

没看懂概率是怎么算的。也没看懂如何通过计算概率得出整除性

AoPSMSE有其它证明。

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 Author| hbghlyj Post time 2023-11-8 19:32
@Mike Spivey的回答
there is no known combinatorial interpretation of the numbers $$A(m,n) = \frac{(2m)! (2n)!}{m! n! (m+n)!}.$$
据此推测1# @Jack D'Aurizio的回答是錯的?

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