AMC 8 · 2022 · #23

Grade 7 counting

Archetype Casework by Position / Vertex Type

combinations-basicsystematic-enumerationspatial-visualization caseworkcomplementary-counting ↑ Prerequisites: combinations-basicsystematic-enumeration

📏 Long solution 💡 4 insights 📊 Diagram

Problem

A $\triangle$ or $\bigcirc$ is placed in each of the nine squares in a $3$ -by- $3$ grid. Shown below is a sample configuration with three $\triangle$ s in a line.

How many configurations will have three $\triangle$ s in a line and three $\bigcirc$ s in a line?

$\textbf{(A) } 39 \qquad \textbf{(B) } 42 \qquad \textbf{(C) } 78 \qquad \textbf{(D) } 84 \qquad \textbf{(E) } 96$

Pick an answer.

(A)

(B)

(C)

(D)

(E)

View mode:

Toolkit + CCSS Solution

Understand

Restated: Each of the $9$ cells of a $3\times 3$ grid is filled with either a triangle ($\triangle$) or a circle ($\bigcirc$). A "line" is any of the $3$ rows, $3$ columns, or $2$ main diagonals — $8$ lines total. Count how many of the $2^9 = 512$ possible fillings simultaneously contain a line of three $\triangle$s AND a line of three $\bigcirc$s.

Givens: A $3 \times 3$ grid with $9$ cells; Each cell holds exactly one of $\triangle$ or $\bigcirc$; A "line" means a row, a column, or a main diagonal ($8$ lines); Required: at least one line is all $\triangle$ AND at least one line is all $\bigcirc$; Answer choices: (A) $39$, (B) $42$, (C) $78$, (D) $84$, (E) $96$

Unknowns: The number of grid fillings that contain both a $\triangle$-line and a $\bigcirc$-line

Understand

Plan

Primary tool: #7 Identify Subproblems

Secondary: #2 Make a Systematic List, #9 Solve an Easier Related Problem

The headline question ("how many grids?") is too big to attack directly, but it cracks open the moment we use Tool #7 (Identify Subproblems) to ask a geometry sub-question first: which pairs of lines can be "all $\triangle$" and "all $\bigcirc$" at the same time? Because the two lines cannot share a cell, only disjoint line pairs survive — and the only disjoint pairs in a $3\times 3$ grid are two-different-rows or two-different-columns. Diagonals are eliminated immediately. That collapses the problem into a much smaller counting task. Tool #9 (Easier Related Problem) handles the remaining piece: count valid fillings under the column case alone, then double by symmetry. Tool #2 (Systematic List) supplies the casework on "how many monochrome columns appear" so nothing is missed or double-counted.

Execute — Answer: D

#7 Identify Subproblems 5.G.B.3 Step 1

Subproblem 1 — which pairs of lines can be the $\triangle$-line and the $\bigcirc$-line?
The two lines must be disjoint (no shared cell).
Check all line-type pairs: any row meets any column in $1$ cell, any row meets either diagonal in $1$ cell, any column meets either diagonal in $1$ cell, and the two diagonals meet at the center.
The only disjoint pairs are two-distinct-rows or two-distinct-columns.

$$\text{disjoint pairs} \in \{\text{row, row}\} \cup \{\text{col, col}\}$$

💡 Sorting the $8$ lines into "can" vs "cannot" coexist is a Grade 5 sorting-by-attribute move that shrinks the problem hugely.

#9 Solve an Easier Related Problem 5.G.B.3 Step 2

By the symmetry of swapping rows and columns, the number of valid grids whose monochrome lines are columns equals the number whose monochrome lines are rows.
Moreover, the two cases never overlap: if a grid has an all-$\triangle$ column and an all-$\bigcirc$ column, then every row already contains both shapes, so no row can be monochrome.
So we count the column case once and multiply by $2$.

$$\text{Total} = 2 \times (\text{configurations with a }\triangle\text{-column AND a }\bigcirc\text{-column})$$

💡 Symmetry between rows and columns is a Grade 5 "properties shared across a category" idea — solve the easier half, then mirror it.

#2 Make a Systematic List 7.SP.C.8 Step 3

Subproblem 2 (Case 1) — exactly one all-$\triangle$ column and exactly one all-$\bigcirc$ column, with the third column mixed.
Pick which of the $3$ columns is all-$\triangle$ ($3$ ways), then which of the remaining $2$ is all-$\bigcirc$ ($2$ ways).
The third column has $3$ cells, each with $2$ choices, giving $2^3 = 8$ fillings; remove the $2$ that are monochrome (those belong to Case 2), leaving $8 - 2 = 6$ valid mixed columns.

$$3 \times 2 \times (2^3 - 2) = 3 \times 2 \times 6 = 36$$

💡 Multiplying independent choices for each column is exactly the Grade 7 "compound events via organized lists" counting principle.

#2 Make a Systematic List 7.SP.C.8 Step 4

Subproblem 2 (Case 2) — all three columns are monochrome, with both shapes present.
Either two columns are $\triangle$ and one is $\bigcirc$ (choose the $\bigcirc$ column in $3$ ways), or two are $\bigcirc$ and one is $\triangle$ ($3$ ways).
Total $3 + 3 = 6$.
Adding Case 1 and Case 2 gives the column total: $36 + 6 = 42$.

$$3 + 3 = 6;\quad 36 + 6 = 42 \text{ (column case)}$$

💡 Splitting the remaining situations into a complete, non-overlapping list and adding is Grade 7 systematic counting.

#9 Solve an Easier Related Problem 7.SP.C.8 Step 5

Apply the symmetry from Step 2: the row case also contributes $42$, and the two cases are disjoint.
So the total number of valid grids is $42 + 42 = 84$, matching choice (D).

$$42 + 42 = 84 \;\Rightarrow\; \textbf{(D)}$$

💡 Adding disjoint cases is the addition principle of counting — Grade 7 compound-event reasoning.

[1] #7 5.G.B.3 Subproblem 1 — which pairs of lines can be the $\triangle$-line and the $\bigcir

[2] #9 5.G.B.3 By the symmetry of swapping rows and columns, the number of valid grids whose mo

[3] #2 7.SP.C.8 Subproblem 2 (Case 1) — exactly one all-$\triangle$ column and exactly one all-$

[4] #2 7.SP.C.8 Subproblem 2 (Case 2) — all three columns are monochrome, with both shapes prese

[5] #9 7.SP.C.8 Apply the symmetry from Step 2: the row case also contributes $42$, and the two

Review

Reasonableness: A quick sanity check on the size of the answer: the total number of $3\times 3$ fillings is $2^9 = 512$. The number of grids with at least one monochrome $\triangle$-line is roughly the same as the number with at least one monochrome $\bigcirc$-line, and most of those do not coincide. Getting $84$ valid grids (about $16\%$ of $512$) is in the right ballpark — definitely below $96$ and well above $39$. The cleaner internal check: the column case alone gives $36 + 6 = 42$, which equals choice (B); doubling for the row case gives $84$, which equals choice (D). The answer choices are deliberately spaced so that someone who forgets to double, or forgets the disjoint diagonal check, lands on $42$ or $78$ — both wrong.

Alternative: Tool #16 (Change Focus / Complement) gives a one-line cross-check on Case 1. Instead of "$8 - 2 = 6$ valid mixed columns," think of the third column as $2^3 = 8$ free fillings, then complement out the $2$ monochrome ones. The total $3 \cdot 2 \cdot 8 = 48$ over-counts grids that actually belong to Case 2 (because a "mixed" column that happens to be monochrome is really Case 2). Subtracting the $6$ Case-2 grids and adding them back as Case 2 gives the same $42$, confirming the column count.

CCSS standards used (min grade 7)

5.G.B.3 Understand that attributes belonging to a category apply to all subcategories (Classifying the $8$ lines (rows, columns, diagonals) into those that can vs cannot be disjoint, and recognizing the row/column symmetry that lets us solve only the column case.)
7.SP.C.8 Find probabilities of compound events using organized lists, tables, and simulation (Counting the column configurations: multiplication principle ($3 \times 2 \times 6 = 36$ in Case 1) and addition principle (Case 1 $+$ Case 2 $= 42$; row $+$ column $= 84$) over a systematic case list.)

⭐ This AMC 8 problem only needs Grade 7 organized-list counting — multiplication and addition of cases — that you already know!

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: D

Review

CCSS standards used (min grade 7)

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