How many matches did these three persons had with each other? We call a tennis tournament cyclic if any two people play with each other. Three players play two plays and then give up and then don't play any other games. If there are $50$ games at all how many matches did these three persons had with each other?

A. $0$
B. $1$
C. $2$
D. $3$
E. $4$

The question seems a bit unclear because it didn't say anything about the other players whether or not they continued the game. Haven't anybody seen a similar question?
 A: I think the correct interpretation of the question is as follows:

A group of people play a tennis tournament. Every two players are scheduled to play each other once. Three of the players each give up and leave after playing two of their scheduled games. All the other players continue and play as many games as they can (i.e. they play all their remaining games except against the three players who've already left). Exactly $50$ games are played in total. How many games took place between players who left early?

With that interpretation, suppose there were $n$ other players (not including the three who left early). We know that all games between these $n$ players took place. That is $n(n-1)/2$ games, i.e. a triangular number. But this number is also $50-k$, where $k$ is the total number of games involving the three players who left early. The maximum possible value of $k$ is $6$ (they play two games each, and all these games are different), and the minimum possible value is $3$ (they each play both the others). So our triangular number is in the range $44\leq n(n-1)/2\leq 47$. There is only one possible triangular number in this range, $45$ (giving $n=10$). This means $k=5$. If the three players play two games each, and a total of $5$ games, that means $1$ game took place between two of the players, and between them they played $4$ games against other players. So the answer is $1$.
