AMC 10 · 2013 · #24
Grade 8 countingCentral 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?
Pick an answer.
AMC 10 2013 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: Two schools each field three players. Call them $A,B,C$ (Central) and $X,Y,Z$ (Northern). Every player must play two games against each opposing player. The whole match runs in six rounds, and in each round three games happen at the same time. Count the number of different schedules for the six rounds.
Givens: Central players $A,B,C$; Northern players $X,Y,Z$; Each cross-team pair (e.g. $A$ vs $X$) plays exactly two games; The match has six rounds, played in order (Round 1, Round 2, ...); Each round has exactly three games, all played simultaneously; Answer choices: (A) $540$, (B) $600$, (C) $720$, (D) $810$, (E) $900$
Unknowns: The number of different ways to schedule all six rounds
Understand
Restated: Two schools each field three players. Call them $A,B,C$ (Central) and $X,Y,Z$ (Northern). Every player must play two games against each opposing player. The whole match runs in six rounds, and in each round three games happen at the same time. Count the number of different schedules for the six rounds.
Givens: Central players $A,B,C$; Northern players $X,Y,Z$; Each cross-team pair (e.g. $A$ vs $X$) plays exactly two games; The match has six rounds, played in order (Round 1, Round 2, ...); Each round has exactly three games, all played simultaneously; Answer choices: (A) $540$, (B) $600$, (C) $720$, (D) $810$, (E) $900$
Plan
Primary tool: #2 Make a Systematic List
Secondary: #4 Introduce a Variable, #15 Organize Information in More Ways, #7 Identify Subproblems
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 $n_\pi$ count how many rounds use permutation $\pi$. Tool #15 (Organize Information in More Ways) turns "each pair plays twice" into a $3\times 3$ 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.
Execute — Answer: E
4.OA.A.3 Step 1 Each round is a one-to-one pairing
- Count the games: each Central player meets each of three Northern players twice, so $3 \times 3 \times 2 = 18$ games.
- Six rounds of three games is also $18$, so no round has an idle player.
- With three simultaneous games and three players per team, each player plays exactly one game per round.
- So a round is a perfect pairing of $\{A,B,C\}$ with $\{X,Y,Z\}$ — a permutation.
- There are $3! = 6$ possible rounds.
💡 Nobody can play two boards at once, so each round just matches every player to one rival — a shuffle of the opponents.
6.EE.B.6 Step 2 Name the counts
- Label the six possible pairings (permutations) $\pi_1,\dots,\pi_6$, where $\pi_1$ is the identity $A\!-\!X, B\!-\!Y, C\!-\!Z$.
- Let $n_i \ge 0$ be how many of the six rounds use pairing $\pi_i$.
- Then $n_1 + \cdots + n_6 = 6$.
- The schedule must make every one of the nine cross-team pairs occur exactly twice.
💡 Instead of tracking a long sequence, just count how often each of the six pairing-types is used.
8.EE.C.8 Step 3 Turn "twice each" into equations
- For each cross-team pair, add up the $n_i$ of the pairings that contain it; that sum must equal $2$.
- For example $A\!-\!X$ sits in the identity $\pi_1$ and in the pairing that only swaps $Y,Z$, giving $n_1 + n_2 = 2$.
- Reading all nine cells of the $3\times 3$ opponent table this way gives nine equations such as $n_1+n_3=2$, $n_1+n_4=2$, $n_4+n_5=2$, and so on.
💡 Each board of the $3\times 3$ grid must be visited exactly twice, and that is just one linear equation per grid cell.
8.EE.C.8 Step 4 Solve the system
- Let $t = n_1$ (the number of identity rounds).
- Substituting through the equations forces every count: $n_2 = n_3 = n_4 = 2 - t$ and $n_5 = n_6 = t$.
- Since every count must be between $0$ and $2$, the only possibilities are $t = 0, 1, 2$.
- So there are exactly three valid families of round-multisets.
💡 One free choice (how many plain identity rounds) locks in everything else, and it can only be $0$, $1$, or $2$.
7.SP.C.8 Step 5 List the three families
- Read off each case.
- $t=1$: every one of the six pairings is used exactly once (six distinct rounds).
- $t=0$: the three "swap" pairings (the transpositions) are each used twice, and no others.
- $t=2$: the identity and the two "rotate" pairings (the $3$-cycles) are each used twice.
- Each family does cover every cross-team pair exactly twice, so 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.8 Step 6 Order the all-distinct family
- Rounds are played in order, so a schedule is an ordered sequence of the six rounds.
- In the $t=1$ family all six rounds are different, so the number of orderings is $6!$.
💡 Six distinct rounds can be slotted into Rounds 1 through 6 in $6!$ orders.
6.NS.B.2 Step 7 Order each doubled family
- In the $t=0$ family (and likewise $t=2$) three pairings each appear twice, so the six rounds form three identical pairs.
- The number of distinct orderings is $6!$ divided by $2!$ for each repeated pairing: $\frac{6!}{2!\,2!\,2!} = \frac{720}{8} = 90$.
- Both doubled families give $90$ each.
💡 Two identical rounds can be swapped without making a new schedule, so divide out a $2!$ for each repeated pairing.
5.OA.A.1 Step 8 Add the families
- Total schedules $= 720$ (all distinct) $+ 90$ ($t=0$) $+ 90$ ($t=2$) $= 900$.
- That is choice (E).
💡 The three legal families never overlap, so just add their ordering-counts.
4.OA.A.3 Count the games: each Central player meets each of three Northern players twice, 6.EE.B.6 Label the six possible pairings (permutations) $\pi_1,\dots,\pi_6$, where $\pi_1 8.EE.C.8 For each cross-team pair, add up the $n_i$ of the pairings that contain it; that 8.EE.C.8 Let $t = n_1$ (the number of identity rounds). Substituting through the equation 7.SP.C.8 Read off each case. $t=1$: every one of the six pairings is used exactly once (s 7.SP.C.8 Rounds are played in order, so a schedule is an ordered sequence of the six roun 6.NS.B.2 In the $t=0$ family (and likewise $t=2$) three pairings each appear twice, so th 5.OA.A.1 Total schedules $= 720$ (all distinct) $+ 90$ ($t=0$) $+ 90$ ($t=2$) $= 900$. Th Review
Reasonableness: The three families are genuinely disjoint (they use different multisets of pairings), so adding is valid, and each was checked to cover all nine cross-team pairs exactly twice. The all-distinct count $6! = 720$ already matches choice (C), which is the trap answer for anyone who forgets the two "doubled" schedules; the correct total adds the two $90$s to reach $900$. A quick lower-bound sanity check: there must be more than $720$ schedules because the six-distinct case is only one of several patterns, and $900$ is the smallest choice above $720$ that our exact count produces. Everything points to $900$.
Alternative: Fix opponents one player at a time (Tool #7, Identify Subproblems). Player $A$'s six opponents can be arranged in $\frac{6!}{2!\,2!\,2!} = 90$ distinct orders over the six rounds. Given $A$'s schedule, count the ways to fill $B$ so that no round forces a conflict for $C$; that inner count works out so that the grand total is $90 \times 10 = 900$. The player-by-player method reaches the same (E) and confirms the permutation-matrix count.
CCSS standards used (min grade 8)
4.OA.A.3Solve multi-step word problems using four operations with whole numbers (Counting the $18$ total games and deducing that each round is a one-to-one pairing with every player active once.)6.EE.B.6Use variables to represent numbers and write expressions to solve problems (Introducing counts $n_i$ for how many rounds use each of the six pairing-types.)8.EE.C.8Analyze and solve pairs of simultaneous linear equations (Writing the nine "each pair appears twice" conditions as a linear system and solving it down to a single parameter $t \in \{0,1,2\}$.)7.SP.C.8Find probabilities of compound events using organized lists, tables, and simulation (Listing the three valid families of round-multisets and counting the $6!$ orderings of the all-distinct family.)6.NS.B.2Fluently divide multi-digit numbers using the standard algorithm (Dividing $6! = 720$ by $2!\,2!\,2! = 8$ to get the $90$ orderings of each doubled family.)5.OA.A.1Use parentheses, brackets, or braces in numerical expressions and evaluate (Adding the three family counts $720 + 90 + 90$ to reach the final total $900$.)
⭐ See 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 $720 + 90 + 90 = 900$.
⭐ See 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 $720 + 90 + 90 = 900$.
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