Paul Kislanko wrote: > The number of full ranked ballots is just the number of permutations of > N alternatives = N! > > If equal ranknigs are allowed, it's N! + 2^N - 1
I assume you mean that there are N! ways to arrange the candidates strictly, and then N - 1 spaces in between that could hold either a > or a =, giving N! * 2^(N - 1). But I think that's an overestimate because it would count both A=B>C and B=A>C. Did you mean something different? -- Rob LeGrand, psephologist [EMAIL PROTECTED] Citizens for Approval Voting http://www.approvalvoting.org/ __________________________________________________ Do You Yahoo!? Tired of spam? Yahoo! Mail has the best spam protection around http://mail.yahoo.com ---- election-methods mailing list - see http://electorama.com/em for list info
