Competition · AMC preparation · step 4 of 4
AMC 8 · 2014 · #16
Grade 7 countingPick an answer.
AMC 8 2014 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
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.
Count the conference pairs
Subproblem 1, conference games: choosing 2 teams from 8 gives 28 distinct pairs.
8 choices for the first team, 7 for the second, divided by 2 because order does not matter — this is the Grade 7 compound-event counting move.
The eight conference teams make 28 different matchup pairs, each pair counted only once.
▸ Why?
Choosing a matchup means picking a first team and then a different second team; that gives 8 × 7 = 56 ordered picks, but each pair is listed twice, so the real count is 56 ÷ 2 = 28.
▸ Why?
There are 8 ways to pick the first team, and for each one there are 7 remaining teams for the second, so the ordered picks are 8 equal groups of 7, which is 8 × 7 = 56.
▸ Why?
Those 56 ordered picks count every matchup twice, once for each order, so the number of true pairs is 56 ÷ 2.
▸ Why?
"A then B" and "B then A" name the very same two teams, so each unordered pair matches exactly one bundle of two ordered picks.
▸ Why?
Counting how many equal bundles of 2 fit inside the 56 ordered picks is exactly what dividing 56 by 2 does, which gives 28.
Double for home and away
Each pair plays 2 games (home and away), so 28 × 2 = 56 conference games.
Equal groups of 2 — Grade 3 multiplication as repeated counting.
3.OA.A.3Identify SubproblemsCount the non-conference games
Subproblem 2, non-conference games: 8 teams × 4 outside games = 32, and the outside opponent means no double-count.
8 equal groups of 4 — straight Grade 3 multiplication, with no double-count to worry about because the opponent is outside.
3.OA.A.3Identify SubproblemsAdd the two counts
Add the two subtotals: 56 + 32 = 88 → (B).
Combining results of two sub-counts into a final total is the Grade 4 multi-step word-problem habit.
4.OA.A.3Identify SubproblemsSplit 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.
- Count the conference pairs
- Double for home and away
- Count the non-conference games
- Add the two counts
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