> -----Original Message----- > From: [EMAIL PROTECTED] > [mailto:[EMAIL PROTECTED] On Behalf Of > Rob LeGrand > Sent: Wednesday, December 14, 2005 3:13 PM > To: Election Methods Mailing List > Subject: Re: [EM] number of possible ranked ballots given N candidates > > 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?
No, it really is 2^N - 1, not 2^(N-1). It's the sum of the ways to select 1,2,...N-1 of N-1 choices for where an = sign can go. ---- election-methods mailing list - see http://electorama.com/em for list info
