回复：回复：Yes, I think C(6,3)*2*2+6!/6=200..

来源: jinjing 于 2012-05-04 08:56:10 [] [旧帖] [给我悄悄话] 阅读数 : (360 bytes)
本帖于 2012-06-08 15:15:07 时间, 由版主 idiot94 编辑
回答: 回复：Yes, I think C(6,3)*2*2+6!/6=200.. 由 Slowguy 于 2012-05-04 07:59:37

Two parts:

There are C(6,3)*C(3,3)=20 cases. For one case,{[a1,a2,a3],[a4,a5,a6]}={[a4,a5,a6],[a1,a2,a3]}.But we have {(a1a2a3,a4a5a6),(a1a2a3,a4a6a5),(a1a3a2,a4a5a6),(a1a3a2,a4a6a5)} 2*2=4 cases for book changes.20*4=80.

a1a2a3 : a1 to a2,a2 to a3 and a3 to a1.

One part:

6!/6=120. round permutation.

80+120=200.

所有跟帖： 

• Still not convinced -Slowguy- ♂ 给 Slowguy 发送悄悄话 (162 bytes) () 05/04/2012 postreply 09:06:41

• You are right, should be C(6,3)/2*2*2 + 6!/6=160. two parts 10 c -jinjing- ♀ 给 jinjing 发送悄悄话 (96 bytes) () 05/04/2012 postreply 10:35:30

• Thanks -Slowguy- ♂ 给 Slowguy 发送悄悄话 (0 bytes) () 05/04/2012 postreply 10:50:22

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