Competition · AMC preparation · step 4 of 4
AMC 8 · 2003 · #16
Grade 4 countingPick an answer.
AMC 8 2003 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
One seat has a special rule (driver) and the other three are unrestricted. Tool #7 (Identify Subproblems) says: handle the constrained seat first, then handle the rest as a separate, simpler subproblem. With the driver picked, the remaining seats become "arrange 3 people in 3 labeled seats," which Tool #13 (Count Smartly) finishes with the multiplication principle: multiply the number of choices at each step. Filling the most-restricted slot first is the standard counting move because it prevents over-counting later.
Count the driver choices
Subproblem 1 — the driver: only Bonnie or Carlo may drive, so the driver seat has exactly 2 choices.
Always start with the slot that has the fewest options — it locks in the hardest piece first.
4.OA.A.3Identify SubproblemsCount the front-seat choices
Subproblem 2 — the front passenger: 3 people remain and anyone may sit there, giving 3 choices.
With no rule blocking anyone, the count is simply "how many people are left."
4.OA.A.3Convert To AlgebraCount the back-left choices
Subproblem 3 — the back-left seat: 2 people are left and either may sit, so 2 choices.
Each filled seat shrinks the leftover pool by one — the choices shrink in lockstep.
4.OA.A.3Convert To AlgebraCount the back-right choices
Subproblem 4 — the back-right seat: just 1 person is left, so the seat is forced — 1 choice.
The last seat is forced once the other three are filled.
4.OA.A.3Convert To AlgebraMultiply the four counts
Combine — the four picks are independent, so multiply the choices: 2 × 3 × 2 × 1 = 12 → (D).
The multiplication principle: when a task splits into independent steps, total arrangements = product of choices at each step.
The total number of valid seating arrangements equals the product of the choices available at each seat, 2 × 3 × 2 × 1.
▸ Why?
Seating the four people breaks into four choices made in order — driver, then front, then back-left, then back-right — and the number of options at each seat (2, then 3, then 2, then 1) is fixed by how many seats are already filled, not by which particular people took the earlier seats.
▸ Why?
Because each seat's option count is set no matter what the earlier picks were, the four choices are independent, and chaining independent choices makes the ways multiply: each of the 2 options for the first seat pairs with all 3 for the next, and so on down the line.
Fill the strictest seat first, then multiply the leftover choices — that simple Grade 4 plan turns this AMC 8 counting problem into 2 × 3 × 2 × 1 = 12.
- Count the driver choices
- Count the front-seat choices
- Count the back-left choices
- Count the back-right choices
- Multiply the four counts
A parent dashboard for the family lives at sensimlab.com.
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