AMC 8 · 2004 · #4
Grade 3 countingProblem
Ms. Hamilton’s eighth-grade class wants to participate in the annual three-person-team basketball tournament.
Lance, Sally, Joy, and Fred are chosen for the team. In how many ways can the three starters be chosen?
Pick an answer.
Toolkit + CCSS Solution
Understand
Restated: From the four players Lance, Sally, Joy, and Fred, three are picked to start. In how many different ways can the three starters be chosen?
Givens: There are $4$ players: Lance, Sally, Joy, Fred; Exactly $3$ players must start each time; Answer choices: (A) $2$, (B) $4$, (C) $6$, (D) $8$, (E) $10$
Unknowns: The number of different three-player starting groups
Understand
Restated: From the four players Lance, Sally, Joy, and Fred, three are picked to start. In how many different ways can the three starters be chosen?
Givens: There are $4$ players: Lance, Sally, Joy, Fred; Exactly $3$ players must start each time; Answer choices: (A) $2$, (B) $4$, (C) $6$, (D) $8$, (E) $10$
Plan
Primary tool: #2 Make a Systematic List
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 $\binom{4}{3}$ 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, $4$.
Execute — Answer: B
2.OA.A.1 Step 1 - Fix an order for the four players so the list cannot repeat or skip a trio.
- Use L (Lance), S (Sally), J (Joy), F (Fred).
💡 Naming the players with single letters keeps the list short and easy to scan.
3.OA.A.3 Step 2 - Write every group of $3$ from $\{L, S, J, F\}$.
- Walk through the players in order and, for each one, leave that player out and group the other three together.
💡 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.
2.OA.A.1 Step 3 Count the trios on the list and match to a choice.
💡 One trio for each player who could sit out, so the count equals the number of players.
2.OA.A.1 Fix an order for the four players so the list cannot repeat or skip a trio. Use 3.OA.A.3 Write every group of $3$ from $\{L, S, J, F\}$. Walk through the players in orde 2.OA.A.1 Count the trios on the list and match to a choice. Review
Reasonableness: The answer must be at least $1$ (one valid trio exists) and at most $4 \times 3 \times 2 = 24$ (the count if order mattered). Since each unordered trio of $3$ players can be written in $3 \times 2 \times 1 = 6$ ordered ways, the unordered count is $24 \div 6 = 4$, matching (B). The trap choice (C) $6$ comes from counting ordered pairs from $4$ players ($4 \times 3 \div 2$) instead of unordered triples; (D) $8$ doubles the answer as if order mattered for $2$ of the slots; (E) $10$ is $\binom{5}{3}$, the right formula with the wrong number of players.
Alternative: Tool #5 (Look for a Pattern) — "choose $3$ to play" is the same as "choose $1$ to sit out." There are $4$ players, so there are $4$ ways to pick one to sit, hence $4$ starting trios. Same answer (B), with no list at all.
CCSS standards used (min grade 3)
2.OA.A.1Use addition and subtraction within 100 to solve one- and two-step word problems (Setting up the player set and counting the final list of trios as a small whole-number total.)3.OA.A.3Use multiplication and division within 100 to solve word problems involving equal groups, arrays, and measurement quantities (Organizing the listing of every $3$-player group from a set of $4$ in a systematic order so each group is counted exactly once.)
⭐ When 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.
⭐ When 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.