let me show U, fn=(n+1)!/2. I'm right,MM is...

Let 1,2,3......,n , i>1+2+......(i-1)

n=1,f1=1, n=2, 23-,32-,3-2, why don't use 1? Cs 1 can put anywhere.

n=3,f3=12    123-,213-,231-,23-1,   132-,312,321-,32-1,   13-2,31-2,3-12,3-21.

We have f(n)=(n+1)!/2 by induction, n=1,2,3 right. n=k right,see n=k+1.

fk=(k+1)!/2, one of permutaions is p1p2p3....pi-p(i+1)...pk.

1 can put in k+2 positions, so we have f(k+1)=(k+2)!/2. We got it.

所有跟帖： 

• Jingling, your answer is obviously wrong. Wxc's proof is beautif -乱弹- ♂ (238 bytes) () 08/04/2011 postreply 10:06:43

• 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