2016 AMC 10B Problems/Problem 22

Problem

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 $A?$

$\textbf{(A)}\ 385 \qquad \textbf{(B)}\ 665 \qquad \textbf{(C)}\ 945 \qquad \textbf{(D)}\ 1140 \qquad \textbf{(E)}\ 1330$

Solution

There are $21$ 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:

$[asy]size(7cm);label("X",(5,5));label("Z",(10,0));label("Y",(0,0));draw((4,4)--(1,1),EndArrow);draw((6,4)--(9,1),EndArrow); label("X",(20,5));label("Z",(25,0));label("Y",(15,0));draw((19,4)--(16,1),EndArrow);draw((16,0)--(24,0),EndArrow);draw((24,1)--(21,4),EndArrow); [/asy]$ But we know that every team beat exactly $10$ other teams, so for each possible $X$ at the head of a fork, there are always exactly $\tbinom{10}2$ choices for $Y$ and $Z$. 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 $\boxed{\textbf{(A)}}$.

Solution 2

Since there are $21$ teams and for each set of three teams there is a cycle, there are a total of $\tbinom{21}3=1330$ cycles of three teams. Because about $1/4$ of the cycles $\{A, B, C\}$ satisfy the conditions of the problems, our answer is close to $1/4*1330=332.5$. Looking at the answer choices, we find that $332.5$ is closer to $385$ than any other answer choices, so our answer is $385$ which is $\boxed{\textbf{(A)}}$.