2016 AMC 12B Problem 20

Each team plays $20$ others, so there are $21$ teams and $\binom{21}{3}=1330$ triples in all

A non-cyclic triple has exactly one team that beats the other two; count those and subtract from the total

Since each team won $10$ and lost $10,$ there are $21$ teams and $\binom{21}{3}=1330$ triples. A triple is not cyclic exactly when one team beats both others. Choosing that team ($21$ ways) and $2$ of the $10$ teams it beat gives $21\cdot\binom{10}{2}=21\cdot45=945$ non-cyclic triples. Thus the cyclic triples number $1330-945=385.$

Thus, the correct answer is A.