Question

A set of teams held a round-robin tournament in which every team played every other team exactly once. Every team won $10$ games and lost $10$ games; there were no ties. How many sets of three teams $\{A,B,C\}$ were there in which $A$ beat $B,$ $B$ beat $C,$ and $C$ beat $A?$

$385$

$665$

$945$

$1140$

$1330$

Solution:

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.

2016 AMC 12B Problem 20

Solution:

Problem 20 in Other Years