AMC 8 · 2022 · #23

Grade 7 counting
combinations-basicsystematic-enumerationspatial-visualization caseworkcomplementary-counting ↑ Prerequisites: combinations-basicsystematic-enumeration
📏 Long solution 💡 4 insights 📊 Diagram
Problem
Each of the 9 cells of a 3× 3 grid is filled with either a 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 2⁹ = 512 possible fillings simultaneously contain a line of three △s AND a line of three bigcircs.

Pick an answer.

(A)
39
(B)
42
(C)
78
(D)
84
(E)
96

AMC 8 2022 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Identify Subproblems

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 △" 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× 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.

1STEP 1

The △-line and bigcirc-line can't share a cell, so the only disjoint options are two different rows or two different columns.

disjoint pairs ∈ {row, row} ∪ {col, col}
2STEP 2

By row–column symmetry the row count equals the column count, and the two cases never overlap, so count the column case and multiply by 2.

Total = 2 × (configurations with a △-column AND a bigcirc-column)
3STEP 3

Case 1 — one all-△ column and one all-bigcirc column: 3 × 2 ways to place them, times 6 mixed fillings of the last column, gives 36.

3 × 2 × (2³ - 2) = 3 × 2 × 6 = 36
4STEP 4

Case 2 — all three columns monochrome with both shapes: 3 + 3 = 6. Adding Case 1 gives the column total 42.

3 + 3 = 6; 36 + 6 = 42 (column case)
5STEP 5

Symmetry gives the row case another 42, and it's disjoint from the column case, so the total is 42 + 42 = 84 — choice (D).

42 + 42 = 84 → (D)
Answer
84
A quick sanity check on the size of the answer: the total number of 3× 3 fillings is 2⁹ = 512. The number of grids with at least one monochrome △-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.
💡Key takeaway

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