2013 AMC 10A Problems/Problem 24

Revision as of 15:13, 8 February 2013 by Countingkg (talk | contribs)

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?

$\textbf{(A)}\ 540\qquad\textbf{(B)}\ 600\qquad\textbf{(C)}\ 720\qquad\textbf{(D)}\ 810\qquad\textbf{(E)}\ 900$


  • Credit to the Math Jam for this solution

Let us label the players of the first team $A$, $B$, and $C$, and those of the second team, $X$, $Y$, and $Z$.

Let us first consider how to organize A's matches, $AX$, $AX$, $AY$, $AY$, $AZ$, and $AZ$. Because we have three duplicates, there are $\frac{6!}{2!*2!*2!} = 90$ ways to organize A's matches.

Now, consider $B$ and $C$. WLOG assume that A's matches were $XXYYZZ$, as we will multiply by $90$ the end anyways, and that, in the first round, $B$ played $Y$ and C played $Z$.

See Also

2013 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 23
Followed by
Problem 25
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