Welcome back,

The answer is n,

We use Balance method:

M1M2.....MnW1W2....Wn    M1W1M2W2M3W3....MnWn   The first is 1,second is 2n-1,average is n,The number of each type both are 2*2n!/(n!*n!)......every pemutation with index n+i,can find dual permutation with index n-i,the number are the same.So the answer is n.

所有跟帖： 

• Correcting: (2n!)!/(n!*n!)types,every type has n!*n! permutation -jinjing- ♀ (0 bytes) () 11/11/2010 postreply 18:06:09

• (2n!)!/(n!*n!)shoud be (2n)!/(n!*n!) -jinjing- ♀ (0 bytes) () 11/12/2010 postreply 08:47:07