## Thinking Mathematically (6th Edition)

Suppose we are using the pairwise comparison method. If there are $n$ candidates, the number of comparisons is $C = \frac{n~(n-1)}{2}$
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 $C = \frac{n~(n-1)}{2}$