A circle is divided into six equal sections. Each section is to be coloured with a single colour so that three sections are red, one is blue, one is green, and one is yellow. Two circles have the same colouring if one can be rotated to match the other. In the diagram, Figure 1 and Figure 2 have the same colouring, while Figure 1 and Figure 3 have different colourings. How many different colourings are there for the circle?
(A) 14
(B) 12
(C) 24
(D) 10
(E) 20

![](https://cdn.mathpix.com/cropped/2024_04_20_46ead6524a8d61e21c51g-005.jpg?height=217&width=201&top_left_y=195&top_left_x=1301)

Figure 1

![](https://cdn.mathpix.com/cropped/2024_04_20_46ead6524a8d61e21c51g-005.jpg?height=238&width=217&top_left_y=198&top_left_x=1515)

Figure 2

![](https://cdn.mathpix.com/cropped/2024_04_20_46ead6524a8d61e21c51g-005.jpg?height=214&width=212&top_left_y=457&top_left_x=1420)

Figure 3
Solution 1

Begin by colouring the section at the top blue.

Since two circles have the same colouring if one can be rotated to match the other, it does not matter which section is coloured blue, so we arbitrarily choose the top section.

![](https://cdn.mathpix.com/cropped/2024_04_20_748fdafcde2a47412608g-009.jpg?height=252&width=247&top_left_y=413&top_left_x=993)

There are now 5 sections which can be coloured green.

After choosing the section to be coloured green, there are 4 sections remaining which can be coloured yellow.

Each of the remaining 3 sections must then be coloured red.

Thus, the total number of different colourings of the circle is $5 \times 4=20$.

## Solution 2

We begin by considering the locations of the three sections coloured red, relative to one another. The three red sections could be adjacent to one another, or exactly two red sections could be adjacent to one another, or no red section could be adjacent to another red section.

We consider each of these 3 cases separately.

Case 1: All three red sections are adjacent to one another

Begin by colouring any three adjacent sections red.

![](https://cdn.mathpix.com/cropped/2024_04_20_748fdafcde2a47412608g-009.jpg?height=252&width=249&top_left_y=1392&top_left_x=995)

Since two circles have the same colouring if one can be rotated to match the other, it does not matter which three adjacent sections are coloured red.

Consider the first section that follows the three red sections as we move clockwise around the circle.

There are 3 choices for the colour of this section: blue, green or yellow.

Continuing to move clockwise to the next section, there are now 2 choices for the colour of this section.

Finally, there is 1 choice for the colour of the final section, and thus there are $3 \times 2 \times 1=6$ different colourings of the circle in which all three red sections are adjacent to one another. These 6 colourings are shown below.
![](https://cdn.mathpix.com/cropped/2024_04_20_748fdafcde2a47412608g-009.jpg?height=248&width=1548&top_left_y=2228&top_left_x=343)

Case 2: Exactly two red sections are adjacent to one another

There are two different possible arrangements in which exactly two red sections are adjacent to one another.

In the first of these, the next two sections that follow the two adjacent red sections as we move clockwise around the circle, are both not red. We call this Case 2a.

In the second of these, the section that follows the two adjacent red sections as we move clockwise around the circle is not red, but the next section is. We call this Case 2b.

The arrangements for Cases 2a and 2b are shown below.
![](https://cdn.mathpix.com/cropped/2024_04_20_748fdafcde2a47412608g-010.jpg?height=282&width=544&top_left_y=604&top_left_x=844)

Notice that the first of these two circles cannot be rotated to match the second.

The number of colourings in Case 2a and in Case $2 \mathrm{~b}$ are each equal to the number of colourings as in Case 1.

That is, there are 3 choices for the first uncoloured section that follows the two red sections as we move clockwise around the circle.

Continuing to move clockwise to the next uncoloured section, there are now 2 choices for the colour of this section.

Finally, there is 1 choice for the colour of the final section, and thus there are $3 \times 2 \times 1=6$ different colourings of the circle in Case 2a as well as in Case 2b.

These 12 colourings are shown below.
![](https://cdn.mathpix.com/cropped/2024_04_20_748fdafcde2a47412608g-010.jpg?height=260&width=1756&top_left_y=1498&top_left_x=258)

Case $2 \mathrm{~b}$
![](https://cdn.mathpix.com/cropped/2024_04_20_748fdafcde2a47412608g-010.jpg?height=260&width=1580&top_left_y=1758&top_left_x=423)

Case 3: No red section is adjacent to another red section Begin by colouring any three non-adjacent sections red.

![](https://cdn.mathpix.com/cropped/2024_04_20_748fdafcde2a47412608g-010.jpg?height=255&width=252&top_left_y=2187&top_left_x=991)

Since two circles have the same colouring if one can be rotated to match the other, it does not matter which three non-adjacent sections are coloured red.

In this case, there are 2 possible colourings as shown below.
![](https://cdn.mathpix.com/cropped/2024_04_20_748fdafcde2a47412608g-011.jpg?height=250&width=544&top_left_y=162&top_left_x=844)

A circle with any other arrangement of the green, yellow and blue sections can be rotated to match one of the two circles above.

The total number of different colourings of the circle is $6+12+2=20$.

ANSWER: (E)