Jingling, your answer is obviously wrong. Wxc's proof is beautif

A subset of valid placements: first put the heaviest on the right, then put others in any order on either side.

The number of such placements is (n-1)! * 2^{n-1}, which is al ready much greater than your answer when n is large enough.

所有跟帖： 

• How are you,Ｍy answer is right is right,though my right side is -jinjing- ♀ (173 bytes) () 08/04/2011 postreply 13:32:34

• my right side is Q's left side, -jinjing- ♀ (0 bytes) () 08/04/2011 postreply 13:35:51

• 唉，怎么说呢。N=3，见内。 -wxcfan123- ♂ (295 bytes) () 08/04/2011 postreply 14:09:00

• Thx,your right,I am sorry ,My answer is for last states.not求整个操作 -jinjing- ♀ (29 bytes) () 08/04/2011 postreply 14:30:48

• My solution, it is not Dct. -jinjing- ♀ (226 bytes) () 08/04/2011 postreply 18:07:45

• Your (n-1)! * 2^{n-1}is not right, should be n!, n=3,first is he -jinjing- ♀ (22 bytes) () 08/04/2011 postreply 14:11:09

• I am sorry ,My answer is for last states.not求整个操作过程的不同方法个数． -jinjing- ♀ (0 bytes) () 08/04/2011 postreply 14:27:34