2013 AMC 10A Problem 24

Label Central’s players $A,B,C$ and Northern’s players $X,Y,Z$. Player $A$’s six-round opponent string contains two each of $X,Y,Z$, so it can be chosen in $\frac{6!}{2!2!2!}=90$ ways. For a fixed $A$-string, say $XXYYZZ$, player $B$’s string must also contain two each of $X,Y,Z$ and cannot match $A$’s opponent in any position. If $B$’s first two entries are $Y,Z$ in either order, then the middle two entries must be $X,Z$ in either order and the last two must be $X,Y$ in either order, giving $2^3=8$ strings. The two remaining possibilities are $YYZZXX$ and $ZZXXYY$, for $10$ total $B$-strings. Once $A$ and $B$ are scheduled, $C$’s schedule is forced. Hence there are $90\cdot10=900$ schedules. Thus, E is the correct answer.