AMC 8 · 2014 · #16

Grade 7 counting
Archetype Arithmetic Expression Decomposition └ Sub-archetype Fraction / Expression Sum Evaluation
combinations-basicmulti-digit-arithmetic identify-subproblems ↑ Prerequisites: combinations-basicmulti-digit-arithmetic
📏 Short solution 💡 2 insights
Problem
The "Middle School Eight" conference has 8 teams. Inside the conference, every team plays every other team twice (1 home game and 1 away game). On top of that, each team also plays 4 games against opponents outside the conference. Count the total number of games in one season that involve at least one Middle School Eight team.

Pick an answer.

(A)
60
(B)
88
(C)
96
(D)
144
(E)
160

AMC 8 2014 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Identify Subproblems

The season has two clearly different kinds of games — conference games (both teams are inside the league) and non-conference games (only one team is inside). Tool #7 (Identify Subproblems) says to split the total into these two independent pieces, count each correctly, and add. Tool #16 (Count Carefully) handles the trap inside the conference subproblem: a game between Team A and Team B is the same game whether we list it as "A vs B" or "B vs A", so we must avoid double-counting the pair. The fact that home and away are distinct games is just a × 2 on top of the unordered pair count.

1STEP 1

Subproblem 1, conference games: choosing 2 teams from 8 gives 28 distinct pairs.

C(8, 2) = 8×72\frac{8 × 7}{2} = 28 pairs
2STEP 2

Each pair plays 2 games (home and away), so 28 × 2 = 56 conference games.

28 × 2 = 56 conference games
3STEP 3

Subproblem 2, non-conference games: 8 teams × 4 outside games = 32, and the outside opponent means no double-count.

8 × 4 = 32 non-conference games
4STEP 4

Add the two subtotals: 56 + 32 = 88 → (B).

56 + 32 = 88 → (B)
Answer
88
Sanity-check the conference count from a single team's view: each team plays the other 7 teams twice, so each team is in 7 × 2 = 14 conference games. Summing over 8 teams gives 8 × 14 = 112, but every conference game has 2 Middle School Eight teams in it, so the actual count is 1122\frac{112}{2} = 56 — matches Step 2. Adding 32 non-conference games gives 88, which is choice (B). The other choices fail simple checks: 160 = 8 × 20 over-counts by treating each team's 20 games independently; 144 doubles conference games without halving; 60 and 96 don't match any clean split.
💡Key takeaway

Split the season into conference games and non-conference games, count each pair only once, then add — basic multiplication plus a careful pair count gets you to 88.

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