AMC 8 · 2003 · #16

Grade 4 counting

Archetype Casework by Position / Vertex Type └ Sub-archetype Vertex or Element Role

permutations-basiccombinations-basicsystematic-enumeration caseworksystematic-enumeration ↑ Prerequisites: multi-digit-arithmeticsystematic-enumeration

📏 Short solution 💡 2 insights

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Problem

Ali, Bonnie, Carlo, and Dianna split into the four seats of a car: 1 driver seat, 1 front passenger seat, and 2 back seats. Only Bonnie and Carlo can drive. Count every possible seating arrangement.

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 8 2003 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Identify Subproblems

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.

1STEP 1

Subproblem 1 — the driver: only Bonnie or Carlo may drive, so the driver seat has exactly 2 choices.

#drivers = 2 (Bonnie or Carlo)

2STEP 2

Subproblem 2 — the front passenger: 3 people remain and anyone may sit there, giving 3 choices.

#front passengers = 3

3STEP 3

Subproblem 3 — the back-left seat: 2 people are left and either may sit, so 2 choices.

#back-left = 2

4STEP 4

Subproblem 4 — the back-right seat: just 1 person is left, so the seat is forced — 1 choice.

#back-right = 1

5STEP 5

Combine — the four picks are independent, so multiply the choices: 2 × 3 × 2 × 1 = 12 → (D).

2 × 3 × 2 × 1 = 12 → (D)

Answer

Without the driver restriction, 4 people in 4 labeled seats would give 4 × 3 × 2 × 1 = 24 arrangements — that is choice (E). The driver rule cuts the valid choices for one seat from 4 down to 2, exactly halving the total: 24 ÷ 2 = 12. The answer 12 matches choice (D). As a second sanity check, the count 12 is between the unrestricted 24 and the smaller choices 2, 4, 6 — it lines up with "only one restriction."

💡Key takeaway

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.

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