AMC 10 · 2025 · #16
Grade 7 countingA circle has been divided into 6 sectors of different sizes. Then 2 of the sectors are painted red, 2 painted green, and 2 painted blue so that no two neighboring sectors are painted the same color. One such coloring is shown below.
How many different colorings are possible?
Pick an answer.
AMC 10 2025 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: A circle is cut into 6 sectors of different sizes, arranged in a ring. Use three colors, red, green, and blue, with each color painting exactly 2 sectors. No two sectors that touch may share a color. Count how many different colorings of the ring obey these rules.
Givens: There are 6 sectors arranged around a circle, so each sector has exactly 2 neighbors.; The sectors have different sizes, so each position is distinct and telling them apart matters.; Each of the 3 colors (red, green, blue) is used on exactly 2 sectors.; No two neighboring sectors may be the same color.
Unknowns: The number of colorings of the 6-sector ring that follow all the rules.
Understand
Restated: A circle is cut into 6 sectors of different sizes, arranged in a ring. Use three colors, red, green, and blue, with each color painting exactly 2 sectors. No two sectors that touch may share a color. Count how many different colorings of the ring obey these rules.
Givens: There are 6 sectors arranged around a circle, so each sector has exactly 2 neighbors.; The sectors have different sizes, so each position is distinct and telling them apart matters.; Each of the 3 colors (red, green, blue) is used on exactly 2 sectors.; No two neighboring sectors may be the same color.
Plan
Primary tool: #2 Make a Systematic List
Secondary: #1 Draw a Diagram, #9 Solve an Easier Related Problem, #7 Identify Subproblems
The question is a 'how many ways' count, so a systematic list is the right engine. To keep the list short, first draw the ring and label the positions, then solve an easier version by fixing the color of one sector and multiplying at the end. The remaining count breaks into a few clean cases based on where the second same-colored sector sits.
Execute — Answer: D
7.SP.C.8 Step 1 Label the ring positions
- Draw the circle and number the sectors 1 through 6 going clockwise.
- Position 1 touches positions 2 and 6, position 2 touches 1 and 3, and so on around the loop, so position 6 touches 5 and 1.
- Because the sectors are different sizes, each numbered spot is its own place, and two colorings that look the same only after turning the circle still count as different.
💡 Naming the spots turns a vague 'circle' into a fixed list you can fill in one position at a time.
3.OA.A.1 Step 2 Fix sector 1, multiply by 3 later
- The three colors all play the same role, so start by counting only the colorings where sector 1 is red.
- Whatever that count is, the same count happens if sector 1 is green and again if sector 1 is blue.
- Since every valid coloring has some single color in sector 1, the three groups do not overlap, so the total is just three equal groups: the sector-1-is-red count multiplied by 3.
💡 Locking one color in place removes repeated work, and the missing colors are recovered by one clean times-3 at the end.
7.SP.C.8 Step 3 Where can the second red go?
- Sector 1 is red, and red is used exactly twice, so there is one more red sector.
- It cannot be a neighbor of sector 1, so it is not sector 2 or sector 6.
- That leaves three cases for the second red: sector 3, sector 4, or sector 5.
- Count each case separately, then add the results.
💡 The one forbidden move, red next to red, cuts the choices down to a short list of positions.
7.SP.C.8 Step 4 Count each case by listing
- Now the four leftover sectors must hold two greens and two blues with no touching pair matching.
- Case A, second red at sector 3: positions 4-5-6 form a chain and sector 2 stands alone, so the chain must alternate (green-blue-green or blue-green-blue) and sector 2 takes the leftover color, giving 2 colorings.
- Case B, second red at sector 4 (straight across): the leftovers split into two separate touching pairs, (2,3) and (5,6); each pair can be green-blue or blue-green, and any combination automatically uses two greens and two blues, giving 2 x 2 = 4 colorings.
- Case C, second red at sector 5: this mirrors Case A, giving 2 colorings.
- Adding the cases gives 2 + 4 + 2 = 8 colorings with sector 1 red.
💡 Once red is pinned down, the rest is a tiny alternating pattern you can just write out and count.
3.OA.A.1 Step 5 Multiply by 3 for all top colors
- There are 8 valid colorings when sector 1 is red.
- The exact same count of 8 holds when sector 1 is green and when it is blue, and these three families never overlap.
- So the full total is three groups of 8.
- That gives 8 x 3 = 24 colorings, which matches answer (D).
💡 Three interchangeable colors mean three copies of the same count, so one multiplication finishes the job.
7.SP.C.8 Draw the circle and number the sectors 1 through 6 going clockwise. Position 1 t 3.OA.A.1 The three colors all play the same role, so start by counting only the colorings 7.SP.C.8 Sector 1 is red, and red is used exactly twice, so there is one more red sector. 7.SP.C.8 Now the four leftover sectors must hold two greens and two blues with no touchin 3.OA.A.1 There are 8 valid colorings when sector 1 is red. The exact same count of 8 hold Review
Reasonableness: A rough upper bound helps: filling positions in order gives 3 choices for the first, then about 2 for each neighbor, far more than 24, and the exact-twice rule plus the no-touch rule should trim it hard, so 24 sits in a believable range. A second check: the three cases 2, 4, 2 are symmetric front-to-back (Case A mirrors Case C), which is what you expect on a symmetric ring, and every case genuinely used two greens and two blues. The total 24 also equals 6 x 4, matching the count another way: 3! = 6 color orderings times 4 structural patterns.
Alternative: Instead of fixing a color, list the two structural shapes directly on the 6-cycle: either all three colors sit on opposite pairs, forcing an alternating pattern ABCABC with 3! = 6 orderings, or exactly one color sits opposite while the other two form mirrored adjacent pairs, giving 3 (which color) x 3 (which opposite pair) x 2 = 18. Then 6 + 18 = 24, confirming (D).
CCSS standards used (min grade 7)
7.SP.C.8Find probabilities of compound events using organized lists, tables, and simulation (Organizing all legal colorings into a systematic case-by-case list and counting the arrangements in each case without missing or repeating any.)3.OA.A.1Interpret products of whole numbers as total number of objects in groups (Combining the three interchangeable top-color families as three equal groups of 8 to get 8 x 3 = 24.)
⭐ Pin one color in place, count the few ways the rest can fit, then multiply back by the colors you set aside.
⭐ Pin one color in place, count the few ways the rest can fit, then multiply back by the colors you set aside.
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