Correcting: (2n!)!/(n!*n!)types,every type has n!*n! permutation
所有跟帖:
• (2n!)!/(n!*n!)shoud be (2n)!/(n!*n!) -jinjing- ♀ (0 bytes) () 11/12/2010 postreply 08:47:07
• (2n!)!/(n!*n!)shoud be (2n)!/(n!*n!) -jinjing- ♀ (0 bytes) () 11/12/2010 postreply 08:47:07
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