[EM] number of possible ranked ballots given N candidates

[Rob LeGrand]

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
rob at approvalvoting.org
Citizens for Approval Voting
http://www.approvalvoting.org/

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