2013 AMC 10A Problem 24

Call Central's players $A, B, C$ and Northern's $X, Y, Z$. Write a game as two letters, like $AX$.

1. What a round looks like. A round has $3$ games and $6$ players, so every player plays exactly once. A round is a way to match $A, B, C$ with $X, Y, Z$, one each. There are $3 \times 2 \times 1 = 6$ ways to do that:

2. Why two groups. The three matchings in Group 1 share no game, and together they contain all $9$ games once. The same is true for Group 2. Also, any matching from Group 1 and any from Group 2 share exactly one game. (For example, the two top rows share $AX$.)
3. The rule for a schedule. Each of the $9$ games is played exactly twice over the $6$ rounds.
4. What that forces. Take any matching $P$ from Group 1 and any $Q$ from Group 2. They share one game. That game is played only in rounds that use $P$ or $Q$ (it is in no other matching). So
$$(\text{times } P \text{ is used}) + (\text{times } Q \text{ is used}) = 2.$$
This holds for every $P$ in Group 1 and every $Q$ in Group 2. So all Group 1 matchings are used the same number of times, say $a$, and all Group 2 matchings are used $2 - a$ times.
5. Three possibilities.
- $a = 1$: all $6$ matchings are used once each. Put them in order: $6 \times 5 \times 4 \times 3 \times 2 \times 1 = 720$ ways.
- $a = 2$: the three Group 1 matchings are each used twice, Group 2 not at all. Six rounds, made of three pairs of identical rounds: $\dfrac{720}{2 \times 2 \times 2} = 90$ orders.
- $a = 0$: the same with Group 2: $90$ orders.
6. Total. $720 + 90 + 90 = 900$.

The answer is $\boxed{\textbf{(E)}\ 900}$.

The key move is to stop thinking about single games and think about whole rounds. Once a round is one of just $6$ matchings, the "each game twice" rule becomes a simple rule about how often each matching is used, and it has only three answers. Counting the orders of rounds is then ordinary arranging, dividing by $2$ for each pair of identical rounds.

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 $\boxed{\textbf{(E)}\ 900}$.