Democratic symmetry (fwd)

[DEMOREP1 at aol.com]

With N choices there are N factorial (!) combinations.

N      N !
2     2
3     6
4    24
5   120
etc.  etc. 

With truncated (short) votes there are additional combinations.

I would suggest that what Prof. Saari finds interesting with the 3 choice 
case (with a mere 6 full vote choices -- i.e. 2 symmetry sets for each first 
choice-- such as N1 BCA and N2 BAC -- to be paired with their respective 
symmetry combinations --- N1 CAB, N1 ABC or N2 ACB, N2 CBA) becomes more than 
a little complex with 4 or more choices.

Even the 3 choice case is rather complex (different N's).

N1 ABC
N2 BCA
N3 CAB

N4 CBA
N5 BAC
N6 ACB

Prof Saari would apparently choose the least amount in the N1-N3 group and 
the least amount in the N4-N6 group to cancel out.

Examples- N3 is the smallest of N1 to N3, N5 is the smallest of N4 to N6-- 
leaving--

N1-N3 ABC
N2-N3 BCA

N4-N5 CBA
N6-N5 ACB

Anybody want to do the resulting Borda-Saari math ?

Very interesting --- 
removing the N3 votes removes a total of 3 x N3 votes
removing the N5 votes removes a total of 3 x N5 votes

This is rather blatant vote stealing.

With 4 or more choices, how many votes would be stolen ?