AMC 8 · 2024 · #25

Grade 7 probability

Archetype Overlap / Complement Counting └ Sub-archetype At-Least-One Complement: 1 − P(None)

probability-basiccombinations-basicsystematic-enumeration complementary-countingsystematic-enumeration ↑ Prerequisites: probability-basiccombinations-basicfraction-arithmetic

📏 Long solution 💡 4 insights 📊 Diagram

Problem

A small airplane has 4 rows of 3 seats each, so 12 seats total. Eight passengers are already seated at random, leaving 4 empty seats. We want the probability that the next-to-board married couple can find 2 adjacent seats in the same row.

Pick an answer.

(A)

$frac{8}{15}$

(B)

$frac{32}{55}$

(C)

$frac{20}{33}$

(D)

$frac{34}{55}$

(E)

$frac{8}{11}$

AMC 8 2024 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Change Focus / Count the Complement

"Probability the couple can sit together" = "probability that among 4 empty seats, at least one adjacent pair lies in the same row." The phrase "at least one" is the textbook trigger for Tool #16 (Complement) — count the opposite event (no two empty seats are adjacent) instead, because it has far fewer cases. To count the opposite, use Tool #2 (Systematic List) on how the 4 empties split across the 4 rows. Tool #1 (Diagram) keeps the 4× 3 grid visible, and Tool #9 (Easier Problem) lets us first work out the tiny one-row subproblem (how many non-adjacent placements fit in a single 3-seat row).

1STEP 1

In one row of 3 seats with no two empties adjacent, the counts for (0,1,2,3) empties are (1,3,1,0) — two empties must be the end seats.

Per-row non-adjacent placements: (0,1,2,3) → (1,3,1,0)

2STEP 2

The denominator is the ways to pick which 4 of 12 seats are empty; order-free, so the combination C(12,4) = 495.

C(12, 4) =

\frac{12 × 11 × 10 × 9}{4 × 3 × 2 × 1}

= 495

3STEP 3

For the opposite event, list how 4 empties split across rows; with ≤2 non-adjacent per row, only (1,1,1,1), (2,1,1,0), (2,2,0,0) qualify.

Partitions: (1,1,1,1), (2,1,1,0), (2,2,0,0)

4STEP 4

Multiply the per-row (1,3,1) counts within each partition: 81, 108, and 6, which sum to 195 non-adjacent arrangements.

81 + 108 + 6 = 195

5STEP 5

The opposite event's probability is $\frac{195}{495}$ ; dividing by the gcd 15 reduces it to $\frac{13}{33}$ — the chance the couple cannot sit together.

P(cannot sit together) =

\frac{195}{495}

\frac{13}{33}

6STEP 6

Apply the complement rule: 1 - $\frac{13}{33}$ = $\frac{20}{33}$ , which matches choice (C); none of the other four fractions equal it.

1 -

\frac{13}{33}

\frac{20}{33}

→ (C)

Answer

\frac{20}{33}

The answer

\frac{20}{33}

≈ 0.606 is just over

\frac{1}{2}

and well below 1 — a sensible probability. With 4 empty seats spread across 4 rows of 3, and each row offering 2 possible adjacent pairs, it is reasonable that the couple usually finds a pair. A quick second check: the number of "good" (couple-can-sit) arrangements is 495 - 195 = 300, and

\frac{300}{495}

\frac{20}{33}

— same answer.

💡Key takeaway

This last AMC 8 problem really only needs the Grade 7 "organized list for compound events" and "complement rule" you already know!

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