The question asks for the total number of unique pairs that can be formed from eleven distinct attitude statements. This is a problem of combinations, as the order of statements in a pair does not matter (e.g., pair A-B is the same as pair B-A).
We need to find the number of ways to choose 2 statements from a set of 11. The formula for combinations is:
$ \binom{n}{k} = \frac{n!}{k!(n-k)!} $
Where:
Applying the formula:
$ \binom{11}{2} = \frac{11!}{2!(11-2)!} $
$ \binom{11}{2} = \frac{11!}{2!9!} $
$ \binom{11}{2} = \frac{11 \times 10 \times 9!}{ (2 \times 1) \times 9!} $
Cancel out $9!$:
$ \binom{11}{2} = \frac{11 \times 10}{2} $
$ \binom{11}{2} = \frac{110}{2} $
$ \binom{11}{2} = 55 $
Therefore, there are 55 possible pairs of attitude statements.