2025 AMC 10B Problem 16

The sectors all have different sizes, so turning the circle gives a different coloring; every sector is its own spot. Number them $1$ to $6$ going around, so $6$ is next to $1$.

1. Split the job in two. Each color is used on exactly two sectors. So a coloring means:
- split the six sectors into three pairs, with no pair being neighbors;
- give the three pairs the three colors, in $3 \times 2 \times 1 = 6$ ways.

1. Count the pairings by sector 1's partner. Sector $1$ cannot pair with its neighbors $2$ or $6$, so its partner is $3$, $4$ or $5$.

Partner 3. Left: $2, 4, 5, 6$. Sector $2$ can pair with $4$, $5$ or $6$.
- $2$ with $4$ leaves $5, 6$: neighbors, not allowed.
- $2$ with $6$ leaves $4, 5$: neighbors, not allowed.
- $2$ with $5$ leaves $4, 6$: fine.

One pairing: $\{1,3\}, \{2,5\}, \{4,6\}$.

Partner 5. This is the mirror image of partner 3. Left: $2, 3, 4, 6$. Sector $2$ can pair with $4$ or $6$.
- $2$ with $6$ leaves $3, 4$: neighbors, not allowed.
- $2$ with $4$ leaves $3, 6$: fine.

One pairing: $\{1,5\}, \{2,4\}, \{3,6\}$.

Partner 4 (the opposite sector). Left: $2, 3, 5, 6$. Sector $2$ can pair with $5$ or $6$.
- $2$ with $5$ leaves $3, 6$: fine.
- $2$ with $6$ leaves $3, 5$: fine.

Two pairings: $\{1,4\}, \{2,5\}, \{3,6\}$ and $\{1,4\}, \{2,6\}, \{3,5\}$.

3. Total. There are $1 + 1 + 2 = 4$ pairings, and each can be colored in $6$ ways:
$$4 \times 6 = 24.$$

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

When every color is used the same number of times, take the colors off first: count the ways to group the spots, then put the colors back with a simple $3 \times 2 \times 1$. Grouping six spots in a circle is small enough to handle by asking "who is sector 1's partner?" and checking each case carefully.

Count all colorings, then remove the bad ones. Placing two reds, two greens and two blues on six sectors can be done in $\frac{6!}{2!\,2!\,2!} = 90$ ways.

• Red pair next to each other: there are $6$ neighboring pairs for red, then the other four sectors take two greens and two blues in $6$ ways, so $36$. The same for green and for blue.
• Red and green both next to each other: red takes a neighboring pair ($6$ ways); the other four sectors are in a row with $3$ neighboring pairs for green, and blue takes what is left. That is $18$ for each of the $3$ pairs of colors.
• All three next to each other: red takes a neighboring pair ($6$ ways); the row of four must split into its first two and last two, with green on either end ($2$ ways). That is $12$.

Adding the three $36$s counts a coloring with two touching pairs twice, so subtract the $18$s. A coloring with all three touching pairs was then added $3$ times and subtracted $3$ times, so add the $12$ back. Bad colorings: $3 \times 36 - 3 \times 18 + 12 = 66$. Good ones: $90 - 66 = 24$.

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 $\boxed{\textbf{(D)}\ 24}$.