AMC10 2016 10B, Problem 22

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 $A?$

Solution: There are $10 \cdot 2+1=21$ teams. Any of the $\tbinom{21}3=1330$ sets of three teams must either be a fork (in which one team beat both the others) or a cycle: [asy]size(7cm);label("A",(5,5));label("C",(10,0));label("B",(0,0));draw((4,4)--(1,1),EndArrow);draw((6,4)--(9,1),EndArrow); label("A",(20,5));label("C",(25,0));label("B",(15,0));draw((19,4)--(16,1),EndArrow);draw((16,0)--(24,0),EndArrow);draw((24,1)--(21,4),EndArrow); [/asy] But we know that every team beat exactly $10$ other teams, so for each possible $A$ at the head of a fork, there are always exactly $\tbinom{10}2$ choices for $B$ and $C$ as $A$ beat exactly 10 teams and we are choosing 2 of them. Therefore there are $21\cdot\tbinom{10}2=945$ forks, and all the rest must be cycles. Thus the answer is $1330-945=385$ which is $\fbox{385}$.

Answer: 385