Dizzy disks!

Probability Level 4

Let a disk be divided in $p$ equal sectors, where $p$ is prime, and colored with $n$ colors. Two colorations which can be deduced one from the other by rotating the disk around its center in some direction, are not considered distinct. We can use the same color in all sectors. In how many ways can this be done? For answering, take $n=4$ and $p=3$

This problem was inspired on some USSR Olympiad problems

The answer is 24.

2 solutions

Maharnab Mitra
May 28, 2014

Chew-Seong Cheong
Jun 4, 2014

The colorations can only involve 1, 2 or 3 colors for the 3 sectors of the disk. We need to find the sum of the numbers of colorations for 1, 2 and 3 colors.

$s = n_{1} + n_{2} + n_{3}$

For 1 color, there are only 4 possible colorations. The number of colarations,

$n_{1} = 4$

For 2 colors, it involves the number of ways or combinations you can choose 2 colors out of 4. And for each combination, there are two colorations with one color occupies one sector and the other occupies two sectors. Therefore,

$n_{2} = 2{ C }_{ 2 }^{ 4 } = 12$

For 3 colors, again for each of the combinations you choose 3 colors out of 4, there are two colorations with two of the colors swap sectors. Therefore,

$n_{3} = 2{ C }_{ 3 }^{ 4 } = 8$

Therefore the answer is

$s = n_{1} + n_{2} + n_{3} = 4 + 12 + 8 = \boxed{24}$

Dizzy disks!

The answer is 24.

2 solutions

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