Question

Central High School is competing against Northern High School in a backgammon match. Each school has three players, and the contest rules require that each player play two games against each of the other school's players. The match takes place in six rounds, with three games played simultaneously in each round. In how many different ways can the match be scheduled?

$540$

$600$

$720$

$810$

$900$

Solution:

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.

2013 AMC 10A Problem 24

Solution:

Problem 24 in Other Years