Various postings have been made regarding combinations of 3 or more choices such as S1 AB S2 S3 AC S4 S5 BC S6 in connection with various methods. With even 3 choices and no truncated votes there are math limits on the numbers. If a choice C is added to the simple S1 A > B and S2 B > A, then there is N1 CAB N2 ACB N3 ABC N4 CBA N5 BCA N6 BAC N1 + N2 + N3 (= S1) A B N4 + N5 + N6 (= S2) N2 + N3 + N6 (= S3) A C N1 + N4 + N5 (= S4) N3 + N5 + N6 (= S5) B C N1 + N2 + N4 (= S6) Put N1 to N6 on the edge of a circle to show the problem. Each of the 6 subparts is in 3 sums. Add a fourth choice and there is major complexity. In other words, all summary head to head results are in doubt unless the raw votes are posted.
