2013 AMC 10A Problem 24

Call Central's players $A, B, C$ and Northern's $X, Y, Z$. There are $9$ possible games, each played twice, for $18$ games in $6$ rounds of $3$. Since $3$ games use all $6$ players, every player plays exactly once per round, so a round is a pairing: a way to match $A, B, C$ with $X, Y, Z$. There are $3! = 6$ pairings.

Split them into two families. Family I: $\{AX, BY, CZ\}$, $\{AY, BZ, CX\}$, $\{AZ, BX, CY\}$; any two of these share no game, and together they cover all $9$ games once. Family II: the other three pairings $\{AX, BZ, CY\}$, $\{AY, BX, CZ\}$, $\{AZ, BY, CX\}$, with the same property. Crucially, each game belongs to exactly one Family I pairing and exactly one Family II pairing.

A schedule is a sequence of $6$ pairings in which every game appears exactly twice. If a game lies in pairing $P$ (Family I) and $Q$ (Family II), then (times $P$ is used) $+$ (times $Q$ is used) $= 2$. Since every $P$ pairs with every $Q$ through some game, all three Family I pairings are used the same number of times $a$, and all Family II pairings $2 - a$ times.

• $a = 1$: each of the $6$ pairings once, in any order: $6! = 720$ schedules.
• $a = 2$: the three Family I pairings twice each: $\frac{6!}{2!\,2!\,2!} = 90$.
• $a = 0$: the three Family II pairings twice each: $90$.

Total: $720 + 90 + 90 = 900$. The answer is $\boxed{\textbf{(E)}\ 900}$.

Recasting each round as a pairing (a permutation of one team) shrinks the search to sequences of $6$ objects drawn from $6$ types. The two families are the even and odd permutations, and the "each game exactly twice" rule becomes a linear constraint on how often each type appears, with only three solutions. Once the multiset of rounds is known, ordering them is a multinomial count.

• Treating the 18 games as freely arrangeable (choosing which 3 go in each round) without enforcing that every player plays exactly once per round.
• Counting only the $720$ schedules that use each pairing once and missing the two "repeat a family" cases.