How many games are played in a round-robin tournament held with n tennis players where each of the players will play against every other player exactly once.
The answer is $\frac{n(n-1)}{2}$. What is wrong with thinking about it this way…
We can line up people in some order $n!$and then say the first two are a pair, the next two are a pair…etc. This over counts by a factor of $\frac{n}{2}! \cdot 2^\frac{n}{2}$ since the order of the pairs doesn't matter, and the order within the pairs doesn't matter.
$$\frac{n!}{\frac{n}{2}! \cdot 2^\frac{n}{2}}$$
Best Answer
The question asks about the number of games over all the rounds, while your proposed answer gives the number of ways to split the players into pairs for one round. They are different questions with different answers. Your answer is fine for the question you are addressing.