A correction to my previous reply. Yes, the derivation I gave considers A=B>C and B=A>C as 2 different cases (I think), but that is still correct if we're talking about the symbolic form of the stored ballots as entered by the voter. I think for the purpose at hand (compression of transmission data) that is appropriate, since the cost of checking for every possible X=Y is significant compared to the cost of storing x X=Y and y Y=X configurations.
Also, from an experimentalists standpoint, there is a difference in changing the = to a > (or >> for such methods) between the two forms, so knowing how many of each form there are is valuable. > -----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? > > -- > 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 > ---- election-methods mailing list - see http://electorama.com/em for list info
