2025 AMC 10B Problem 16

The sectors have different sizes, so all six are distinguishable and no colouring is identified with a rotation of itself. Only the cyclic adjacency matters, so label the sectors $1, 2, \dots, 6$ in order around the circle, with $6$ next to $1$.

Each colour is used on exactly two sectors, so a colouring amounts to two independent choices:

1. split the six sectors into three pairs, no pair being two neighbours;
2. give the three pairs the three colours, in $3! = 6$ ways.

Count the pairings. Sector $1$ must be paired with $3$, $4$ or $5$.

Partner $3$. Sectors $2, 4, 5, 6$ remain. Sector $2$ can take $4$, $5$ or $6$: taking $4$ leaves the neighbours $5, 6$; taking $6$ leaves the neighbours $4, 5$; taking $5$ leaves $4, 6$, which are not neighbours. One pairing, $\{1,3\},\{2,5\},\{4,6\}$.

Partner $5$. The mirror image of the previous case. Sector $2$ must take $4$ (taking $6$ would leave the neighbours $3, 4$). One pairing, $\{1,5\},\{2,4\},\{3,6\}$.

Partner $4$. Sectors $2, 3, 5, 6$ remain, and sector $2$ can take $5$ or $6$. Taking $5$ leaves $3, 6$; taking $6$ leaves $3, 5$. Both work, so two pairings, $\{1,4\},\{2,5\},\{3,6\}$ and $\{1,4\},\{2,6\},\{3,5\}$.

That is $1 + 1 + 2 = 4$ pairings, and each gives $6$ colourings:
$$4 \cdot 6 = 24.$$

The answer is $\boxed{\textbf{(D)}\ 24}$.

When every colour is used the same number of times, it pays to strip the colours off first and count the underlying partition, then put the colours back with a clean factorial. What remains is a small structural question, "how many ways can six points on a cycle be matched in non-adjacent pairs", which yields to one round of casework on a single sector's partner. The factor $3!$ at the end is exact because the three pairs are distinguishable as sets, so no two colourings are ever counted twice.

Complementary counting. There are $\frac{6!}{2!\,2!\,2!} = 90$ ways to place the multiset of colours on the six sectors. Let $A_{\text{red}}$ be the event that the two red sectors are neighbours, and similarly for green and blue. There are $6$ neighbouring pairs on the circle, so $|A_{\text{red}}| = 6 \cdot \frac{4!}{2!\,2!} = 36$. For two colours at once, choosing a neighbouring pair for red leaves four sectors in a row, which contain $3$ neighbouring pairs, so $|A_{\text{red}} \cap A_{\text{green}}| = 6 \cdot 3 = 18$. For all three, the remaining row of four must split into its two end pairs, in $1$ way, with $2$ ways to name them, so the triple intersection is $6 \cdot 2 = 12$. Inclusion-exclusion gives $3(36) - 3(18) + 12 = 66$ bad colourings, and $90 - 66 = 24$.

• Finding only three of the four valid pairings, typically missing one of the two that go with a pair of opposite sectors, which gives 3 * 3! = 18 at choice (C).
• Assuming the two sectors of a colour must be diametrically opposite, which counts only the single pairing $\{1,4\},\{2,5\},\{3,6\}$ and gives $6$.
• Dividing by $6$ for rotations; the sectors have different sizes precisely so that rotated colourings are different colourings.