AMC 8 · 2018 · #11

Grade 7 probabilitycounting

Archetype Casework by Position / Vertex Type

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

📏 Medium solution 💡 3 insights 📊 Diagram

Problem

Abby, Bridget, and four of their classmates will be seated in two rows of three for a group picture, as shown.
\begin{eqnarray*} \text{X}&\quad\text{X}\quad&\text{X} \ \text{X}&\quad\text{X}\quad&\text{X} \end{eqnarray*}
If the seating positions are assigned randomly, what is the probability that Abby and Bridget are adjacent to each other in the same row or the same column?

$\textbf{(A) } \frac{1}{3} \qquad \textbf{(B) } \frac{2}{5} \qquad \textbf{(C) } \frac{7}{15} \qquad \textbf{(D) } \frac{1}{2} \qquad \textbf{(E) } \frac{2}{3}$

Pick an answer.

(A)

$\frac{1}{3}$

(B)

$\frac{2}{5}$

(C)

$\frac{7}{15}$

(D)

$\frac{1}{2}$

(E)

$\frac{2}{3}$

View mode:

Toolkit + CCSS Solution

Understand

Restated: Six children — Abby, Bridget, and four classmates — are seated at random in a $2 \times 3$ 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)?

Givens: $6$ seats arranged in $2$ rows of $3$; $6$ children are seated uniformly at random (every seating is equally likely); Abby and Bridget are two specific children among the six; Answer choices: (A) $\tfrac{1}{3}$, (B) $\tfrac{2}{5}$, (C) $\tfrac{7}{15}$, (D) $\tfrac{1}{2}$, (E) $\tfrac{2}{3}$

Unknowns: The probability that Abby and Bridget occupy adjacent seats (same row OR same column)

Understand

Plan

Primary tool: #1 Draw a Diagram

Secondary: #2 Make a Systematic List, #7 Identify Subproblems

The problem is fundamentally spatial — "adjacent in a $2 \times 3$ 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 $\binom{6}{2}$.

Execute — Answer: C

#1 Draw a Diagram 2.G.A.2 Step 1

Draw the seating chart and label the seats.
Use a $2 \times 3$ grid of dots with labels so we can point at specific seats when counting.
The top row is seats $1, 2, 3$ and the bottom row is seats $4, 5, 6$.

$$\begin{array}{ccc} 1 & 2 & 3 \\ 4 & 5 & 6 \end{array}$$

💡 Partitioning a rectangle into rows and columns of same-size seats is a Grade 2 array idea — exactly what this picture is.

#2 Make a Systematic List 7.SP.C.8 Step 2

Count the total number of ways to choose which two seats Abby and Bridget occupy.
Because we only care about adjacency (not who sits where within the pair), we count unordered pairs of seats out of $6$.

$$\binom{6}{2} = \dfrac{6 \times 5}{2 \times 1} = 15$$

💡 Listing all unordered $2$-seat selections from $6$ seats is the "organized list" step inside probability counting (Grade 7).

#7 Identify Subproblems 1.OA.A.1 Step 3

Count the horizontal (same-row) adjacent pairs by walking through each row.
In the top row, $\{1,2\}$ and $\{2,3\}$ share edges.
In the bottom row, $\{4,5\}$ and $\{5,6\}$ share edges.
That is $2 + 2 = 4$ horizontal pairs.

$$\text{Horizontal pairs} = 2 + 2 = 4$$

💡 Adding two same-size groups (top row + bottom row) is a Grade 1 addition word problem in disguise.

#7 Identify Subproblems 1.OA.A.1 Step 4

Count the vertical (same-column) adjacent pairs.
Each column has just one front-back pair: $\{1,4\}$, $\{2,5\}$, $\{3,6\}$.
That is $3$ vertical pairs.

$$\text{Vertical pairs} = 1 + 1 + 1 = 3$$

💡 Three columns, each contributing one pair — count by ones, the simplest Grade 1 addition.

#7 Identify Subproblems 1.OA.A.1 Step 5

Add the two cases to get the total favorable count.
Horizontal and vertical adjacencies are disjoint (no pair is both in the same row and the same column), so they add without overlap.

$$\text{Favorable pairs} = 4 + 3 = 7$$

💡 Combining two non-overlapping groups by adding is the Grade 1 "put-together" model.

#1 Draw a Diagram 7.SP.C.7 Step 6

Compute the probability as the ratio of favorable pairs to total pairs.
This is exactly the definition of probability for equally-likely outcomes.

$$P(\text{adjacent}) = \dfrac{7}{15} \;\Rightarrow\; \textbf{(C)}$$

💡 Probability as (favorable outcomes) $\div$ (total equally-likely outcomes) is the Grade 7 probability-model definition.

[1] #1 2.G.A.2 Draw the seating chart and label the seats. Use a $2 \times 3$ grid of dots with

[2] #2 7.SP.C.8 Count the total number of ways to choose which two seats Abby and Bridget occupy

[3] #7 1.OA.A.1 Count the horizontal (same-row) adjacent pairs by walking through each row. In t

[4] #7 1.OA.A.1 Count the vertical (same-column) adjacent pairs. Each column has just one front-

[5] #7 1.OA.A.1 Add the two cases to get the total favorable count. Horizontal and vertical adja

[6] #1 7.SP.C.7 Compute the probability as the ratio of favorable pairs to total pairs. This is

Review

Reasonableness: Sanity-check the magnitudes. There are $15$ pairs of seats and the answer is $\tfrac{7}{15} \approx 0.467$, just under one-half — that feels right because nearly half of the seat pairs in a small $2 \times 3$ 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 $\tfrac{4 \cdot 2 + 2 \cdot 3}{6} = \tfrac{14}{6} = \tfrac{7}{3}$ average neighbors; the probability is then $\tfrac{7/3}{5} = \tfrac{7}{15}$, confirming the answer.

Alternative: Tool #3 (Eliminate Possibilities) works directly on the answer choices: the denominator must come from a count out of $\binom{6}{2} = 15$, so any reduced fraction must have $15$, $5$, $3$, or $1$ as denominator. Choice (A) $\tfrac{1}{3} = \tfrac{5}{15}$, (B) $\tfrac{2}{5} = \tfrac{6}{15}$, (C) $\tfrac{7}{15}$, (D) $\tfrac{1}{2}$ ($2$ does not divide $15$, eliminate), (E) $\tfrac{2}{3} = \tfrac{10}{15}$. Choice (D) is immediately out. Then we only need to count favorable pairs — exactly $7$ — to lock in (C).

CCSS standards used (min grade 7)

2.G.A.2 Partition a rectangle into rows and columns of same-size squares (Drawing and labeling the $2 \times 3$ seating grid so adjacency can be reasoned about visually.)
1.OA.A.1 Solve addition and subtraction word problems within 20 (Adding the counts of horizontal pairs ($2 + 2 = 4$), vertical pairs ($1 + 1 + 1 = 3$), and combining them ($4 + 3 = 7$).)
7.SP.C.8 Find probabilities of compound events using organized lists, tables, and simulation (Counting the total $\binom{6}{2} = 15$ unordered seat pairs for the sample space.)
7.SP.C.7 Develop probability models and use them to find probabilities of events (Forming the probability as the ratio of favorable seat pairs to total seat pairs: $\tfrac{7}{15}$.)

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

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: C

Review

CCSS standards used (min grade 7)

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