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 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 ListFor 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.
3.OA.A.3Make A Systematic ListFour 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.