[EM] number of possible ranked ballots given N candidates

[Paul Kislanko]

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: 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?
> 
> --
> Rob LeGrand, psephologist
> rob at approvalvoting.org
> 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
>