Competition · AMC preparation · step 4 of 4

AMC 8 · 2018 · #11

Grade 7 probabilitycounting

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

probability-basiccombinations-basicsystematic-enumerationspatial-visualization caseworksystematic-enumeration ↑ Prerequisites: fraction-arithmeticcombinations-basic

📏 Medium solution 💡 3 insights 📊 Diagram

Problem

Six children — Abby, Bridget, and four classmates — are seated at random in a 2 × 3 grid of seats (2 rows, 3 seats per row). What is the probability that Abby and Bridget end up in seats that are adjacent, meaning they share an edge horizontally (same row, side-by-side) or vertically (same column, one in front of the other)?

Pick an answer.

(A)

$\frac{1}{3}$

(B)

$\frac{2}{5}$

(C)

$\frac{7}{15}$

(D)

$\frac{1}{2}$

(E)

$\frac{2}{3}$

AMC 8 2018 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Draw a Diagram

The problem is fundamentally spatial — "adjacent in a 2 × 3 grid" — so Tool #1 (Draw a Diagram) is the natural entry point: sketch the six seats and physically mark which pairs share an edge. Once the picture is drawn, Tool #2 (Systematic List) lets us count adjacent pairs without missing or doubling any, by walking through them in a fixed order. Tool #7 (Identify Subproblems) splits the count into two clean cases — horizontal (same-row) adjacencies and vertical (same-column) adjacencies — so each case is easy to handle on its own. The probability itself is then the favorable pair count divided by the total pair count C(6, 2).

1STEP 1

Draw and number the seats

Draw the 2 × 3 grid and label the seats: top row 1, 2, 3 and bottom row 4, 5, 6, so we can point at each seat when counting.

1 & 2 & 3 ; 4 & 5 & 6

2STEP 2

Count all seat pairs

We care only which two seats the pair takes, so count unordered pairs from the 6 seats — 15 total pairs.

C(6, 2) = (6 × 5)/(2 × 1) = 15

3STEP 3

Count the side-by-side pairs

Walk each row: the top has {1,2} and {2,3}, the bottom has {4,5} and {5,6} — 4 horizontal pairs.

Horizontal pairs = 2 + 2 = 4

4STEP 4

Count the up-down pairs

Each column has one front-back pair — {1,4}, {2,5}, {3,6} — so 3 vertical pairs.

Vertical pairs = 1 + 1 + 1 = 3

5STEP 5

Add the two counts

Same-row and same-column adjacencies never overlap, so add them: 4 + 3 = 7 favorable pairs.

Favorable pairs = 4 + 3 = 7

6STEP 6

Form the probability

Divide favorable by total for equally-likely outcomes: $\frac{7}{15}$ → (C).

P(adjacent) = 7/15 → (C)

Answer

7/15

Sanity-check the magnitudes. There are 15 pairs of seats and the answer is

\frac{7}{15}

≈ 0.467, just under one-half — that feels right because nearly half of the seat pairs in a small 2 × 3 grid share an edge. Spot-check by picking any seat: a corner seat (like seat 1) is adjacent to 2 others (2 and 4); an edge-middle seat (like seat 2) is adjacent to 3 others (1, 3, 5). Averaging the four corners (2 neighbors each) and two middles (3 neighbors each) gives

\frac{4 · 2 + 2 · 3}{6}

\frac{14}{6}

\frac{7}{3}

average neighbors; the probability is then

\frac{\frac{7}{3}}{5}

\frac{7}{15}

, confirming the answer.

💡Key takeaway

This AMC 8 problem only needs Grade 7 probability — favorable outcomes divided by total outcomes — that you already know!

Draw and number the seats
Count all seat pairs
Count the side-by-side pairs
Count the up-down pairs
Add the two counts
Form the probability

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