AMC10 2013 A

Answer (E): Call the players from Central $A$, $B$, and $C$, and call the players from Northern $X$, $Y$, and $Z$. Represent the schedule for each Central player by a string of length six consisting of two each of $X$, $Y$, and $Z$. There are $\binom{6}{2}\binom{4}{2}\binom{2}{2}=90$ possible strings for player $A$. Assume without loss of generality that the string is $XXYYZZ$. Player $B$’s schedule must be a string with no $X$’s in the first two positions, no $Y$’s in the next two, and no $Z$’s in the last two. If $B$’s string begins with a $Y$ and a $Z$ in either order, the next two letters must be an $X$ and a $Z$, and the last two must be an $X$ and a $Y$. Because each pair can be ordered in one of two ways, there are $2^3=8$ such strings. If $B$’s string begins with $YY$ or $ZZ$, it must be $YYZZXX$ or $ZZXXYY$, respectively. Hence there are $10$ possible schedules for $B$ for each of the $90$ schedules for $A$, and $C$’s schedule is then determined. The total number of possible schedules is $900$.