Subjects combinatorics

Painted Cubes 8Fa9B6

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1. **Problem statement:** We start with an unpainted wooden cube. We paint exactly three faces: one red, one blue, and one green. The other three faces remain unpainted. We want to find how many distinct cubes can be formed, where two cubes are considered the same if one can be rotated or flipped to become the other. 2. **Key idea:** The problem is about counting distinct colorings of the cube's faces under the cube's rotational symmetries. The cube has 6 faces, and we paint exactly 3 faces with distinct colors (red, blue, green), and leave 3 faces unpainted. 3. **Total ways without symmetry:** Choose which 3 faces to paint out of 6: $$\binom{6}{3} = 20$$ ways. 4. **Assign colors to the chosen 3 faces:** The 3 chosen faces are painted with 3 distinct colors, so number of ways to assign colors is $$3! = 6$$. 5. **Total colorings without considering symmetry:** $$20 \times 6 = 120$$. 6. **Accounting for cube symmetries:** The cube's rotation group has 24 elements. We use Burnside's Lemma to count distinct colorings: Burnside's Lemma formula: $$\text{Number of distinct colorings} = \frac{1}{|G|} \sum_{g \in G} \text{Fix}(g)$$ where $|G|=24$ and $\text{Fix}(g)$ is the number of colorings fixed by symmetry $g$. 7. **Analyze each type of rotation:** - Identity rotation fixes all 120 colorings. - Rotations about axes through centers of opposite faces (90°, 180°, 270°): - 90° or 270° rotations cycle 4 faces, so no coloring with 3 distinct painted faces fixed. - 180° rotations swap pairs of faces; no coloring fixed because colors differ. - Rotations about axes through midpoints of opposite edges (180°): - These swap pairs of faces; no coloring fixed. - Rotations about axes through opposite vertices (120°, 240°): - These cycle 3 faces and 3 other faces; no coloring fixed because colors differ. 8. **Conclusion:** Only the identity rotation fixes all 120 colorings. All other rotations fix 0 colorings. 9. **Apply Burnside's Lemma:** $$\frac{1}{24} (120 + 0 + 0 + \cdots + 0) = \frac{120}{24} = 5$$ **Final answer:** There are **5** distinct cubes that can be formed under the given conditions.