We can use the 'table' idea to derive the formula for the total number of head-to-head matchups involving n candidates.
Since this table has N rows and N columns, there are N^{2} empty cells below.
Candidate 1 | Candidate 2 | . . . . . | Candidate N | |
Candidate 1 | . . . . . | |||
Candidate 2 | . . . . . | |||
: | : | : | : | : |
Candidate N | : |
But, there are N cells on the main diagonal (candidate 1 vs. candidate 1, candidate 2 vs. candidate 2, etc.) which should be eliminated since they do not represent real matchups. If we take away those N cells (from the original N^{2} cells, we are left with N^{2} -N cells (see the table below).
Candidate 1 | Candidate 2 | . . . . . | Candidate N | |
Candidate 1 | XXXXX | . . . . . | ||
Candidate 2 | XXXXX | . . . . . | ||
: | : | : | . . . . . | : |
Candidate N | . . . . . | XXXX |
However, half of the remaining cells are just duplicates of each other (candidate 1 vs. candidate 2 and candidate 2 vs. candidate 1, etc.). So dividing the N^{2} -N cells (from above) by two, we get (N^{2} -N)/2 cells. If we factor out an N, we get an even better way to write this formula: N(N -1)/2 .
Thus, there are N(N -1)/2 total head-to-head matchups in an N candidate election.