Ooooh now I see what you meant. No, then 120 is not the answer. 120 is the number of all possible 3 element (team) subsets of the set with 10 teams. This is found using combinations without repetition, google “10 choose 3”. But your problem has a restriction, so we have to subtract the number of combinations that DO NOT satisfy this restriction, aka we need to find the “bad” combinations (for example, {1,2,3} is a bad combination, beacuse 1 and 2 played a game).

There are 5 pairs that played a game. Let’s assume that we chose a combination {1,2,x}, that is a bad combination because 1 and 2 played a game. The question is, how many of these combinations are there? Well, we can put any of the remaining elements as x, so there are 8 of those combinations. We repeat this process for the remaining bad pairs, and there are 5 of them, so, the number of all possible bad combinations is 5*8=40, so the final answer is 120-40=80.

This can be generalized, if you changed the grouping size from 3 to n, then the total number of combinations would be 10 choose n. The number of bad combinations would be 5*X, where X is the number of ways we can choose the remaining n-2 teams (which gets a little more complicated but nothing too much).