pigeonhole principle related problem

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I'm given the problem: In a tournament which 18 teams participate, a team being matched with another in a round don’t match again in the follwoing (later) rounds. After 8 rounds prove that there are 3 teams not being matched with each other.

I don't know where to start from. Can anyone help me to aproach to the solution of this problem

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The number of matches needed for all teams to have faced each other exactly once is $$\binom{18}2=153$$ Since each team should only be able to play against one team at a time, the number of distinct matches played in total is $$9\cdot 8 = 72$$ This leaves $153-72=81$ unplayed matches.
What does your statement about "$3$ teams not being matched with each other" mean?