AMC 10 · 2013 · #24
Grade 8 countingPick an answer.
AMC 10 2013 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
This is a "how many ways" count, so Tool #2 (Make a Systematic List) sets the target: list every legal way the six rounds can be built. The clean move is to see each round as one permutation of {X,Y,Z} against {A,B,C} (each player plays once per round), then use Tool #4 (Introduce a Variable): let n_π count how many rounds use permutation π. Tool #15 (Organize Information in More Ways) turns "each pair plays twice" into a 3 × 3 table of counts, which becomes a small system of linear equations. Solving it (Tool #7, Identify Subproblems: first find the valid round-multisets, then count orderings of each) gives just three families, and the answer is the total number of ordered schedules they generate.
Each round is a one-to-one pairing
18 games exactly fill six rounds of three, so everyone plays once per round: a round pairs A,B,C with X,Y,Z — 3! = 6 possible rounds.
Nobody can play two boards at once, so each round just matches every player to one rival — a shuffle of the opponents.
Nobody plays two boards at once, so each round simply matches every player to one rival.
▸ Why?
Each player takes exactly one opponent, so players and opponents pair off without leftovers.
▸ Why?
Each round's pairing is chosen apart from the others, so the schedule counts multiply across rounds.
Name the counts
Name the six pairings π₁,…,π₆ (π₁ the identity) and let n_i count the rounds using π_i, so n₁ + … + n₆ = 6.
Instead of tracking a long sequence, just count how often each of the six pairing-types is used.
6.EE.B.6Introduce A VariableTurn "twice each" into equations
Each cross-team pair sits in exactly two pairings, so the nine cells of the 3 × 3 table give nine equations like n₁ + n₂ = 2.
Each board of the 3 × 3 grid must be visited exactly twice, and that is just one linear equation per grid cell.
8.EE.C.8Organize Information In More WaysSolve the system
With t = n₁, the equations force n₂ = n₃ = n₄ = 2 - t and n₅ = n₆ = t, and each count lies in 0..2, so t = 0, 1, 2.
One free choice (how many plain identity rounds) locks in everything else, and it can only be 0, 1, or 2.
8.EE.C.8Identify SubproblemsList the three families
t=1 uses all six pairings once; t=0 doubles the three swaps; t=2 doubles the identity and the two rotations. All three are legal.
Either use all six pairings once, or double up on the three swaps, or double up on the identity-plus-two-rotations.
7.SP.C.8Make A Systematic ListOrder the all-distinct family
Rounds are played in order, so the six all-different rounds of the t=1 family can be arranged in 6! = 720 ways.
Six distinct rounds can be slotted into Rounds 1 through 6 in 6! orders.
7.SP.C.8Make A Systematic ListOrder each doubled family
A doubled family repeats three pairings, so swapping twins makes nothing new: 6!/2! 2! 2! = 90 orderings each.
Two identical rounds can be swapped without making a new schedule, so divide out a 2! for each repeated pairing.
6.NS.B.2Identify SubproblemsAdd the families
The three families never overlap, so add them: 720 + 90 + 90 = 900, choice (E).
The three legal families never overlap, so just add their ordering-counts.
5.OA.A.1Identify SubproblemsSee each round as one full shuffle pairing the two teams, require every match-up to happen twice, and you find only three schedule patterns: six different shuffles (720 orders) or a doubled pattern of two kinds (90 each), giving 720 + 90 + 90 = 900.
- Each round is a one-to-one pairing
- Name the counts
- Turn "twice each" into equations
- Solve the system
- List the three families
- Order the all-distinct family
- Order each doubled family
- Add the families