Pigeonhole Problem (Scrimmage Match) Suppose you have 10 players on a team, and you want to arrange matches (5 against 5) such that no pair of players are always on the same team. How many matches do you need? I think Pigeonhole principle is used here, but I'm not sure why my pigeonholes would be? Not sure if it helps but if I consider the 10 players {1, 2, ..., 10} and let the first match be {1,2,3,4,5} vs {6, 7, 8, 9, 10}. Then I would need 4 more matches to avoid having 2 of the same pairs on all teams: 1. {1,7,8,9,10} vs {6,2,3,4,5} 2. {1,2,8,9,10} vs {6,7,3,4,5} 3. {1,2,3,9,10} vs {6,7,8,4,5} 4. {1,2,3,4,10} vs {6,7,8,9,5} So, I would need a total of 5 matches at minimum? Not sure if this is helpful. How should I proceed?

It looks like you want to have no pair of players on the same team in every game. After the first game you have two groups of five who have played together. If you split them as evenly as possible you will have two groups of three and two pairs after the second game. In the third game two out of each group of three must still be together, but you might succeed in four games. Your schedule does not break up pairs as quickly as possible. How about $$1,2,3,4,5 \text{ vs } 6,7,8,9,10\\\ 1,3,5,7,9 \text{ vs } 2,4,6,8,10\\\ 1,3,4,7,10 \text{ vs } 2,5,6,8,9\\\ 1,2,4,5,6 \text{ vs } 3,7,8,9,10$$ After the third game the only pairs we needed to split were $1,3$ and $6,8$