2016 AMC 10B Problem 22

Each team played $10 + 10 = 20$ games, one against every other team, so there are $21$ teams and $\binom{21}{3} = 1330$ sets of three teams.

Look at the three games inside any set of three teams. Either the results form a cycle ($A$ beats $B$, $B$ beats $C$, $C$ beats $A$), or they do not; and if they do not, then some team won both of its games in the set. Exactly one team can do so (two teams cannot both beat each other), so each non-cyclic triple is identified by its dominant team together with two of that team's victims.

Count non-cyclic triples by dominant team: each of the $21$ teams beat $10$ others, giving $\binom{10}{2} = 45$ pairs of victims, for
$$21 \cdot 45 = 945$$
non-cyclic triples in total.

The cyclic triples are what remain:
$$1330 - 945 = 385.$$

The answer is $\boxed{\textbf{(A)}\ 385}$.

Cycles are hard to count directly but their complement has a clean structure: a non-cyclic triple has a unique "winner of the triple," and choosing that winner plus two of its victims is a bijection with such triples. The uniform record ($10$ wins each) makes the count $\binom{10}{2}$ the same for every team. This is a standard tournament argument: transitive triples $= \sum_i \binom{w_i}{2}$, cyclic triples $= \binom{n}{3} - \sum_i \binom{w_i}{2}$.

• Reporting the number of transitive triples, $945$ (choice C), or the total number of triples, $1330$ (choice E), instead of their difference.
• Using $20$ teams (forgetting that a team does not play itself), which gives $\binom{20}{3} = 1140$ (choice D) and a wrong count.