Counting

Let p be a permutation of some m-element set. Suppose p can be expressed as the product of c(p) disjoint cycles. If, for some integer n (at least m) we sum sgn(p)n^c(p) over all of the permutations of the m-element set, we get (n choose m)*m!. Prove this -- if you look at it a bit, it seems pretty intuitive, but writing it clearly is tricky.

Originally posted by royalchicken
Let p be a permutation of some m-element set. Suppose p can be expressed as the product of c(p) disjoint cycles. If, for some integer n (at least m) we sum sgn(p)n^c(p) over all of the permutations of the m-element set, we get (n choose m)*m!. Prove this -- if you look at it a bit, it seems pretty intuitive, but writing it clearly is tricky.

I didnt understand what you said.

Originally posted by royalchicken
[b If you look at it a bit, it seems pretty intuitive, but writing it clearly is tricky.[/b]

Yeah, I mean, I solved this pathetically easy problem, but I mean...writing it is just too hard for me.....Or was it reading it.....

After 3 years of colloege algebra I understand the question right up until you start talking about sgn. WTF is that. I think you might want to ask your lecturer this.

Originally posted by Virak
After 3 years of colloege algebra I understand the question right up until you start talking about sgn. WTF is that. I think you might want to ask your lecturer this.

It's the signature of the permutation; I'm not asking for help with coursework; this is not a coursework problem. I've proved it, but given that it's something I thought of and not something someone told me, my proof could be erroneous and it may not be true at all. I'll look at my argument again, although I'm fairly sure it's legit.