(2n!)!/(n!n!)shoud be (2n)!/(n!n!)

jinjing · 2010-11-12 08:47:07Z

(2n!)!/(n!*n!)shoud be (2n)!/(n!*n!)簡介

來源: jinjing 於 2010-11-12 08:47:07 [檔案] [舊帖] [給我悄悄話] 閱讀數 : (0 bytes)

回答: Correcting: (2n!)!/(n!*n!)types,every type has n!*n! permutation 由 jinjing 於 2010-11-11 18:06:09

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