For N candidates,
possible ballots = sum( f(N,m) )
The summation is over all integer values of m from 1 to N, inclusive. The function
f(N,m) represents the number of ways N objects can be put into m slots without
leaving any empty slots. The recursive definition of f(N,m) is:
f(N,1) = 1
f(N,m) = N! for N=m
f(N,m) = m( f(N-1,m) + f(N-1,m-1) ) for all other cases
Richard
Rob LeGrand wrote:
[EMAIL PROTECTED]">Martin wrote:Ok, if there are n candidats then there are n! ways to vote a fully
ranked ballot, and {int(e x n!)} ways to vote a truncated ballot, or
{int((e-1) x (n!) - 1)} ways if you count votes like A>B>C(>D) as
equivalent to A>B>C>D. All this I've found out by reading around
websites and such.
However, I can't seem to find anywhere which says how many ways there
are to vote a ranked ballot which allows draws in arbitrary places, and
I can't see any way to work it out. Any maths/stats people here know
what the answer is, or where I might find out?
I haven't been able to figure out a simple general formula, but here are my
results for up to 6 candidates:
candidates: 1 2 3 4 5 6
possible ballots: 1 3 13 75 541 4683
=====
Rob LeGrand
[EMAIL PROTECTED]
http://www.aggies.org/honky98/
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