I have no idea what going on in this question, nor how to approach it. From messing around with this question I found 14 pairs (i.e. (1,2),(2,1) is considered one pair) that I can add to the original set. But surely it cannot be right? Because then my options will be astronomically large..
Recall that a partition defines an equivalence relation and vice versa. We may look instead at the possible partitions.
Being told that it contains $(6,6)$ is irrelevant as every equivalence relation will have this per reflexivity.
We are told that $7,4,2$ must be in the same equivalence class (since (7,4) and (4,2) are required elements). We are told that $5$ must be in a different equivalence class than $4$ (since (5,4) is a required nonelement). We are told that $3$ must be in a different equivalence class than both $7$ and $5$ (since (7,3) and (3,5) are required nonelements). We are told that $1$ must be in the same equivalence class as $3$ (since (1,3) is a required element).
So... we know that there are at least three equivalence classes, $\{1,3\},\{2,4,7\},\{5\}$. The only question is then for the remaining number $6$... whether it goes into one of these already existing equivalence classes or if it forms its own equivalence class by itself. There are merely $4$ options.