AMC 10 · 2006 · #7

Grade 7 counting
Archetype Branching State Enumeration └ Sub-archetype Ordered Placement Count
fundamental-counting-principlepermutations-basicsystematic-enumeration systematic-enumeration ↑ Prerequisites: fundamental-counting-principle
📏 Medium solution 💡 2 insights
Problem
Four family members take four car seats, and only the two adults may drive. Count the seating arrangements.

Pick an answer.

(A)
4
(B)
12
(C)
16
(D)
24
(E)
48
How to solve
Strategy Make a Systematic List

Four seats and four people is small enough to count in one clean sweep, but only if the seats are pinned down first. Sketch the car so the four seats are named, then fill them in an order that puts the restricted seat first: once the driver is chosen, nothing is restricted any more and the rest is a plain arrangement of three people in three seats. Then count a second time by a different split — grouping arrangements by which pair sits in front — so the total is confirmed by a route that never uses the first one's logic.

1STEP 1

Name the four seats

Naming the seats lets them be filled one at a time.

Seats: D (driver), F (front passenger), B₁, B₂
2STEP 2

Fill the restricted seat first

Filling the restricted seat first keeps every later count fixed.

2_D · 3_F · 2_B₁ · 1_B₂
3STEP 3

Multiply the choices

Multiplying the choices gives 12.

2 · 3 · 2 · 1 = 2 · 6 = 12
4STEP 4

Recount by who sits in front

Recounting by who sits in front confirms 12, choice (D).

2 · 2_two adults in front + 4 · 1 · 2_one adult, one child in front + 0_two children in front = 4 + 8 = 12
Answer
12
The answer must sit strictly between two easy bounds. With no rule at all there are 4 × 3 × 2 × 1 = 24 arrangements, so the answer is below 24, killing (D) 24 and (E) 48 — and 48 is more arrangements than there are ways to seat four people at all. At the other end, just the arrangements with the two adults in front already number 4, and there are clearly more than those, so (A) 4 is too small. That leaves (B) 12 and (C) 16, and 16 fails a divisibility check: swapping Mr. and Mrs. Lopez's seats pairs up the valid arrangements with the valid arrangements in which the other adult drives, so the total splits evenly into two equal halves of 6, and 6 + 6 = 12.
💡Key takeaway

When one spot has a rule and the rest do not, fill the ruled spot first — after that, the leftover people just line up freely.

  • Name the four seats
  • Fill the restricted seat first
  • Multiply the choices
  • Recount by who sits in front

More like this

Same archetype — closest grade level first.