2016 AMC 12B Problem 20

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?

Show solution

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.