Competition · AMC preparation · step 4 of 4
AMC 8 · 2004 · #4
Grade 3 countingPick an answer.
AMC 8 2004 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
With only 4 players and groups of 3, the whole answer fits on one short list. Tool #2 (Make a Systematic List) is the kid-friendly way to count: pick a fixed order for the players, then write each possible trio once. This avoids Tool #13 (Algebra) and the C(4, 3) formula — neither is needed when the entire answer space has at most a handful of outcomes. A clean shortcut also drops out of the list: choosing 3 to keep is the same as choosing 1 to leave out, so the count must equal the number of players, 4.
Fix an order for the players
Fix an order for the four players and give them short labels: L (Lance), S (Sally), J (Joy), F (Fred).
Naming the players with single letters keeps the list short and easy to scan.
2.OA.A.1Make A Systematic ListList every trio
For each player, leave that one out and group the other three: {S,J,F}, {L,J,F}, {L,S,F}, {L,S,J}.
Each row is named by who is missing: leave out L, then S, then J, then F. The systematic order guarantees no trio is missed or repeated.
The number of different three-player starting groups equals the number of players, because each group is fixed the moment you name the single player left out.
▸ Why?
Choosing three players to start is the very same act as choosing the one player to leave out, so every group carries a unique tag: the player who sits.
▸ Why?
The four players split cleanly into the one left out and the three who start, with nobody uncounted and nobody in both parts, so naming who sits out fixes exactly which three remain.
▸ Why?
Every group pairs with exactly one left-out player and every player left out yields exactly one group, so the groups and the players match up one for one and are equal in number.
Count the trios
Four groups appear, one per player left out — that matches choice (B).
One trio for each player who could sit out, so the count equals the number of players.
2.OA.A.1Make A Systematic ListWhen there are only a few players, writing out every possible group beats any formula — and noticing that picking 3 to play is the same as picking 1 to sit makes the count obvious. With that shift, this AMC 8 problem becomes a Grade 3 systematic-counting exercise.
- Fix an order for the players
- List every trio
- Count the trios
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