The most a team could play is 12, if they played all of the other six teams twice each.

The fewest possible is 6, if they played each of the other teams once. So on the face of it it looks as if the seven different possible numbers of games are 6,7,8,9,10,11,12.

However if a team played each other team once there can’t be team that played every team twice, as that would include the first team, leading to a contradiction.

So there are two scenarios regarding number of games played: either 6,7,8,9,10,11 or 7,8,9,10,11,12. If you were expecting seven different possibilities, reread the question and find that only the other six team’s numbers of games have to be different, and Nuneaton Predators number of games can be a repeat of one of those. That is the key to finding the number of games played by the Predators.

Let’s just consider the 7,8,9,10,11,12 scenario. One team, let’s call it A, must play every team twice, to become a team with 12 games. Another team, say B, plays A twice and every other team once, totalling 7 games. Team C plays A twice and B once, and every other team twice to be the team with 11 games. Team D plays teams A and C twice each, and team B and every other team only once, to total 8. Team E plays B and D once each and every other team twice, making 10. Teams F and G have played A, C and E twice, and B and D once, making 8 so far. If they play each other once, they will both be on 9 games, but if they play each other twice they will both be on 10 games, along with team E. In that second situation, no team has played exactly 9 times, so that can’t be the correct situation. Those two teams play 9 games each, and one them must be Nuneaton Predators.

If you do the same thing considering the 6,7,8,9,10,11 scenario, you will again find that the repeated number of games is 9, and so one of those teams must be Nuneaton Predators.

Nuneaton Predators played nine times.