I want you to explain this to me step by step.---**Question Stem:** In the figure below, 3 of the 6 disks are to be painted blue, 2 are to be painted red, and 1 is to be painted green. Two paintings that can be obtained from one another by a rotation or a reflection of the entire figure are considered the same. How many different paintings are possible? **Chart/Diagram Description:** * **Type:** Geometric figure consisting of a stack of circular disks. * **Main Elements:** * **Shapes:** The figure consists of 6 identical outlined circles (disks). * **Relative Position:** The circles are arranged in a triangular formation, resembling a stack or pyramid: * The top row has 1 circle. * The middle row has 2 circles, positioned below and supporting the top circle. * The bottom row has 3 circles, positioned below and supporting the middle row. * Adjacent circles are touching. The arrangement implies a compact, equilateral triangular shape for the overall structure of the disks' centers.

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We have six disks arranged in a triangular formation, with one disk on top, two in the middle row, and three on the bottom. We need to paint three disks blue, two disks red, and one disk green. The key constraint is that rotations and reflections of the entire figure are considered the same painting. This makes it a problem involving Burnside's lemma from group theory, which helps us count distinct arrangements under symmetry operations. The triangular arrangement of disks has six symmetries that form the dihedral group D3. There are three rotations: the identity transformation, a 120-degree clockwise rotation, and a 240-degree clockwise rotation. There are also three reflections: one through vertex 1 to the midpoint of the opposite edge, one through vertex 2, and one through vertex 3. Each symmetry operation maps the triangle to itself, but may permute the positions of the disks. To solve this problem, we use Burnside's lemma, which counts distinct objects under group actions. The formula states that the number of distinct colorings equals one over the size of the group, times the sum of fixed colorings for each group element. Our symmetry group has 6 elements, so we need to count how many colorings with 3 blue, 2 red, and 1 green disk remain unchanged under each symmetry operation. For the identity transformation, all 60 possible colorings are fixed since nothing moves. We calculate this as 6 choose 3 for blue positions, times 3 choose 2 for red positions, times 1 choose 1 for green, giving us 20 times 3 times 1 equals 60. For the 120 and 240 degree rotations, positions form cycles: position 1 stays fixed, positions 2 and 3 form a 2-cycle, and positions 4, 5, and 6 form a 3-cycle. For a coloring to remain unchanged, positions in the same cycle must have the same color, but this makes it impossible to satisfy our constraint of exactly 3 blue, 2 red, and 1 green disk. Therefore, both rotations have zero fixed colorings. For each reflection, one position stays fixed on the reflection line, while the other positions form pairs that must have the same color. We systematically count the valid colorings for each case. If the fixed position is blue, we need 2 more blues and 2 reds and 1 green among the pairs, which is impossible. If the fixed position is red, we need 3 blues, 1 red, and 1 green among the pairs, giving us 3 valid arrangements. If the fixed position is green, we need 3 blues and 2 reds among the pairs, also giving us 3 valid arrangements. So each reflection has 6 fixed colorings. Applying Burnside's lemma: one-sixth times the sum of 60 plus 0 plus 0 plus 6 plus 6 plus 6 equals 78 over 6, which equals 13 distinct paintings.