Answer
Suppose we are using the pairwise comparison method. If there are $n$ candidates, the number of comparisons is $C = \frac{n~(n-1)}{2}$
Work Step by Step
With the pairwise comparison method, each candidate is compared with every other candidate. If there are $n$ candidates, then each candidate must be compared with each of the other $n-1$ candidates.
We can determine the number of comparisons which are necessary when using the pairwise comparison method. We can number the candidates from 1 to $n$.
Candidate 1 must be compared with candidate 2, candidate 3, and so on, up to candidate $n$. This requires $n-1$ comparisons.
Candidate 2 must be compared with candidate 3, candidate 4, and so on, up to candidate $n$. This requires $n-2$ comparisons.
We can count the comparisons like this until candidate $n-1$ is compared with candidate $n$. This requires 1 comparison.
To find the total number $C$ of required comparisons, we need to sum all of these comparisons.
$C$ = (n-1) + (n-2) + ... + 2 + 1
$C = \frac{n~(n-1)}{2}$
