Jobst already provided the correct answer, which is part of the mathematical literature. No need to publish yours.
> -----Original Message----- > From: [EMAIL PROTECTED] > [mailto:[EMAIL PROTECTED] On Behalf Of > Warren Smith > Sent: Wednesday, December 14, 2005 7:02 PM > To: [email protected] > Subject: [EM] counting ballots > > >Kislanko: > 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 > > If truncation is allowed it is approximately N! * e > > And if both truncation AND equal rankings are allowed it's > approximately N! > * e + 2^N - 1 > > > --the statements containing "2^N" are false. > > The true answers are in one of the chapters of my (as yet unpublished) > book. Anyhow, letting E(N) denote the answer, there are > various direct and indirect > expressions for E(N). > E(N) = #ballots for N candidates with equal rankings allowed. > Truncation forbidden. > E(0)=E(1)=1, E(2)=3 E(3)=13 E(4)=75 E(5)=541 > Amazingly, E(N) is the nearest integer to > N! / (2*ln(2)^(N+1)) when N=0,1,...,15,16 > but this is false when N=17. It is true asymptitically for N large. > > Generating function: > sum E(N) * x^N / N! = 1/(2-exp(x)) > N>=0 > > > > I also have the exact result if truncatio is allowed too, but > it is messier. > wds > ---- > election-methods mailing list - see http://electorama.com/em > for list info > ---- election-methods mailing list - see http://electorama.com/em for list info
