A math question

Six friends will exchange books in their book club. Each friend has one book to give
to a friend, and will receive one book from a different friend. (No two friends trade
books with each other.) In how many ways can the books be exchanged?

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• Can wait jinjing, but not an easy one -Slowguy- ♂ (41 bytes) () 05/03/2012 postreply 21:56:32

• I think C(6,3)*2*2+6!/6=200.. -jinjing- ♀ (0 bytes) () 05/04/2012 postreply 06:17:44

• 回复：I think C(6,3)*2*2+6!/6=200.. -slowguy- ♂ (50 bytes) () 05/04/2012 postreply 06:25:34

• Yes, I think C(6,3)*2*2+6!/6=200.. -jinjing- ♀ (320 bytes) () 05/04/2012 postreply 07:07:28

• 回复：Yes, I think C(6,3)*2*2+6!/6=200.. -Slowguy- ♂ (871 bytes) () 05/04/2012 postreply 07:59:37

• 回复：回复：Yes, I think C(6,3)*2*2+6!/6=200.. -jinjing- ♀ (360 bytes) () 05/04/2012 postreply 08:56:10

• Still not convinced -Slowguy- ♂ (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- ♀ (96 bytes) () 05/04/2012 postreply 10:35:30

• Thanks -Slowguy- ♂ (0 bytes) () 05/04/2012 postreply 10:50:22

• Yes, Slowguy is right -万斤油- ♂ (0 bytes) () 05/04/2012 postreply 11:15:16

• If two people can exchange each other.the answer is 265 cases. -jinjing- ♀ (0 bytes) () 05/04/2012 postreply 10:57:58

• If the total people is 7, what is the number of 3+4 part? Is any -wxcfan123- ♂ (0 bytes) () 05/05/2012 postreply 08:08:56

• 回复：If the total people is 7, what is the number of 3+4 part? Is -jinjing- ♀ (120 bytes) () 05/05/2012 postreply 10:04:16

• thanks. So the general solution should be -wxcfan123- ♂ (416 bytes) () 05/06/2012 postreply 08:28:32

• F(N) = SUM_(n1 + n2 + ... nk=N, ni >=3 ) {[C(N, n1)C(N-n1,n2).. -jinjing- ♀ (66 bytes) () 05/06/2012 postreply 09:37:01

• ... {[C(N, n1)C(N-n1,n2)..C(N-n1-n2-...n(k-1,nk)/D(n1,n2m,,,nk)] -jinjing- ♀ (0 bytes) () 05/06/2012 postreply 09:41:59