<html><head></head><body>I came up with a recursive formula and tested it against Rob's numbers. It seems to work.<br>
For N candidates,<br>
<br>
possible ballots = sum( f(N,m) )<br>
<br>
The summation is over all integer values of m from 1 to N, inclusive. The function<br>
f(N,m) represents the number of ways N objects can be put into m slots without<br>
leaving any empty slots. The recursive definition of f(N,m) is:<br>
<br>
f(N,1) = 1<br>
f(N,m) = N! for N=m<br>
f(N,m) = m( f(N-1,m) + f(N-1,m-1) ) for all other cases<br>
<br>
Richard<br>
<br>
<br>
Rob LeGrand wrote:<br>
<blockquote type="cite" cite="mid:20010430203123.53265.qmail@web11005.mail.yahoo.com"><pre wrap="">Martin wrote:<br></pre>
<blockquote type="cite"><pre wrap="">Ok, if there are n candidats then there are n! ways to vote a fully <br>ranked ballot, and {int(e x n!)} ways to vote a truncated ballot, or <br>{int((e-1) x (n!) - 1)} ways if you count votes like A>B>C(>D) as <br>equivalent to A>B>C>D. All this I've found out by reading around <br>websites and such.<br><br>However, I can't seem to find anywhere which says how many ways there <br>are to vote a ranked ballot which allows draws in arbitrary places, and <br>I can't see any way to work it out. Any maths/stats people here know <br>what the answer is, or where I might find out?<br></pre></blockquote>
<pre wrap=""><!----><br>I haven't been able to figure out a simple general formula, but here are my<br>results for up to 6 candidates:<br><br>candidates: 1 2 3 4 5 6<br>possible ballots: 1 3 13 75 541 4683<br><br>=====<br>Rob LeGrand<br><a class="moz-txt-link-abbreviated" href="mailto:honky98@aggies.org">honky98@aggies.org</a><br><a class="moz-txt-link-freetext" href="http://www.aggies.org/honky98/">http://www.aggies.org/honky98/</a><br><br>__________________________________________________<br>Do You Yahoo!?<br>Yahoo! Auctions - buy the things you want at great prices<br><a class="moz-txt-link-freetext" href="http://auctions.yahoo.com/">http://auctions.yahoo.com/</a><br><br><br><br></pre>
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