AMC 8 · 2022 · #23
Grade 7 countingProblem
A △ or ◯ is placed in each of the nine squares in a 3-by-3 grid. Shown below is a sample configuration with three △s in a line.
How many configurations will have three △s in a line and three ◯s in a line?
Pick an answer.
AMC 8 2022 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
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
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) $96$
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
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.
💡 Sorting the $8$ lines into "can" vs "cannot" coexist is a Grade 5 sorting-by-attribute move that shrinks the problem hugely.
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 $2$.
💡 Symmetry between rows and columns is a Grade 5 "properties shared across a category" idea — solve the easier half, then mirror it.
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.
💡 Multiplying independent choices for each column is exactly the Grade 7 "compound events via organized lists" counting principle.
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: $36 + 6 = 42$.
💡 Splitting the remaining situations into a complete, non-overlapping list and adding is Grade 7 systematic counting.
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).
💡 Adding disjoint cases is the addition principle of counting — Grade 7 compound-event reasoning.
5.G.B.3 Subproblem 1 — which pairs of lines can be the $\triangle$-line and the $\bigcir 5.G.B.3 By the symmetry of swapping rows and columns, the number of valid grids whose mo 7.SP.C.8 Subproblem 2 (Case 1) — exactly one all-$\triangle$ column and exactly one all-$ 7.SP.C.8 Subproblem 2 (Case 2) — all three columns are monochrome, with both shapes prese 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.3Understand 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.8Find 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!
⭐ This AMC 8 problem only needs Grade 7 organized-list counting — multiplication and addition of cases — that you already know!
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