$20$ tennis players are scheduled to play exactly $14$ two-person games among themselves, with each player playing at least one game. Prove there must be a set of $6$ games with $12$ distinct players.
To set it up, I let $G$ be a graph on $20$ vertices (players) and $14$ edges (games). Two vertices are adjacent if and only if they play each other in a game. Note that since each player plays a game, each vertex has degree at least $1$.
Using $\displaystyle{\sum_{i=1}^{20} deg(v_i) = 28} $, we can show that at least $12$ vertices have degree $1$. Also, we can show that the largest degree any vertex can have is $8$.
I believe we should show there is a matching of size $6$ on $G$.
Is this a good method? If so, any suggestion on how to show this would be helpful. Thank you.
Suppose you have atmost $5$ games with distinct players. Let the set of these distinct players be $S$. Then the remaining $10$ players can't play any game among themselves because if they did, you would have $6$ games with distinct players. Thus each of these $10$ players must play a game with someone from $S$. This will make $5+10=15$ games which is a contradiction.