[EM] number of possible ranked ballots given N candidates
Paul Kislanko
kislanko at airmail.net
Wed Dec 14 13:29:19 PST 2005
> -----Original Message-----
> From: election-methods-bounces at electorama.com
> [mailto:election-methods-bounces at electorama.com] 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.
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