AMC 8 · 2003 · #16
Easy mode Grade 4Problem
Ali, Bonnie, Carlo, and Dianna are driving to a theme park in one car. The car has 4 seats: 1 driver's seat, 1 front passenger seat, and 2 back seats.
There is one rule. Only Bonnie and Carlo know how to drive. So the driver's seat must be one of them.
The other three seats can be filled by anyone left over.
How many different ways can the four of them sit in the car?
Pick an answer.
AMC 8 2003 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: 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.
Givens: Four people: Ali, Bonnie, Carlo, Dianna; Four labeled seats: driver, front passenger, back-left, back-right; Only Bonnie and Carlo are allowed in the driver seat; The other three seats have no restriction; Answer choices: (A) $2$, (B) $4$, (C) $6$, (D) $12$, (E) $24$
Unknowns: The total number of valid seating arrangements
Understand
Restated: 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.
Givens: Four people: Ali, Bonnie, Carlo, Dianna; Four labeled seats: driver, front passenger, back-left, back-right; Only Bonnie and Carlo are allowed in the driver seat; The other three seats have no restriction; Answer choices: (A) $2$, (B) $4$, (C) $6$, (D) $12$, (E) $24$
Plan
Primary tool: #7 Identify Subproblems
Secondary: #13 Count Smartly
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.
Execute — Answer: D
4.OA.A.3 Step 1 - Subproblem 1 — pick the driver.
- The rule restricts the driver seat to Bonnie or Carlo.
- That gives exactly $2$ choices for who drives.
💡 Always start with the slot that has the fewest options — it locks in the hardest piece first.
4.OA.A.3 Step 2 - Subproblem 2 — pick the front passenger.
- After one person sits in the driver seat, $3$ people remain.
- Any of them can take the front passenger seat, so there are $3$ choices.
💡 With no rule blocking anyone, the count is simply "how many people are left."
4.OA.A.3 Step 3 - Subproblem 3 — pick the back-left passenger.
- Two people remain after the driver and front passenger are seated.
- Either of them can take the back-left seat, so $2$ choices.
💡 Each filled seat shrinks the leftover pool by one — the choices shrink in lockstep.
4.OA.A.3 Step 4 - Subproblem 4 — pick the back-right passenger.
- Only $1$ person is left, so that person takes the last seat.
- $1$ choice.
💡 The last seat is forced once the other three are filled.
4.OA.A.3 Step 5 - Combine the subproblems.
- The four choices are independent (each pick is a separate decision), so multiply the option counts together.
💡 The multiplication principle: when a task splits into independent steps, total arrangements $=$ product of choices at each step.
4.OA.A.3 Subproblem 1 — pick the driver. The rule restricts the driver seat to Bonnie or 4.OA.A.3 Subproblem 2 — pick the front passenger. After one person sits in the driver sea 4.OA.A.3 Subproblem 3 — pick the back-left passenger. Two people remain after the driver 4.OA.A.3 Subproblem 4 — pick the back-right passenger. Only $1$ person is left, so that p 4.OA.A.3 Combine the subproblems. The four choices are independent (each pick is a separa Review
Reasonableness: Without the driver restriction, $4$ people in $4$ labeled seats would give $4 \times 3 \times 2 \times 1 = 24$ arrangements — that is choice (E). The driver rule cuts the valid choices for one seat from $4$ down to $2$, exactly halving the total: $24 \div 2 = 12$. 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."
Alternative: Tool #2 (Make a Systematic List): fix the driver as Bonnie. The remaining $3$ people (Ali, Carlo, Dianna) fill the other $3$ seats in $3! = 6$ ways. Repeat with Carlo as driver: another $6$ ways with the remaining (Ali, Bonnie, Dianna). Total $= 6 + 6 = 12$, again giving (D).
CCSS standards used (min grade 4)
4.OA.A.3Solve multistep word problems using the four operations (Splitting the seating into four sequential choices ($2$, $3$, $2$, $1$) and multiplying them to get the total $2 \times 3 \times 2 \times 1 = 12$.)
⭐ Fill the strictest seat first, then multiply the leftover choices — that simple Grade 4 plan turns this AMC 8 counting problem into $2 \times 3 \times 2 \times 1 = 12$.
⭐ Fill the strictest seat first, then multiply the leftover choices — that simple Grade 4 plan turns this AMC 8 counting problem into $2 \times 3 \times 2 \times 1 = 12$.
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