RE: Condorcet pairs on the ballot

[DEMOREP1]

With N candidates, there are (N x (N-1))/2 pairings. 

N    Pairings
2   1
3   3
4   6
5   10
6   15
It is obviously easier to do N rankings than to do the (N x (N-1))/2
pairings. 
Example- With N=5, it is obviously easier for a voter to put 5 rankings on a
ballot rather than vote on all 10 pairings.