Arrangement of 15 balls including 3 each of 5 different colors in a triangle
Fifteen balls including 3 each of 5 different colors are arranged in a triangle as shown. How many ways can this be done if rotations are allowed?
I was thinking the answer should be $15!/(3*(3!)^5)$ as we can arrange 15 balls in 15 positions in 15! ways. Then since there are 3 balls of 5 different colors each, we divide it by $(3!)^5$ and then divide by 3 as rotation is allowed.
But this is not correct as if 6 balls are needed to arrange like this where there are 3 balls each of 2 different colors, by this logic, the answer should be $6!/3!3!3$ which is not an integer.
$\endgroup$ 32 Answers
$\begingroup$This is almost exactly the correct answer. The error lies in rotationally symmetric arrangements.
The way I prefer to think about it is to say that a "picture" is a representation of an arrangement of $15$ balls where we draw all $15$ of them in a triangle (something like the image in the diagram, but colored differently). We can count the pictures easily: there are $\frac{15!}{3!^5}$ pictures, because we just choose the colors of the balls. There are no rotations to worry about.
Almost every arrangement of $15$ balls has exactly $3$ pictures to go with it. However, some arrangements are rotationally symmetric, and in that case, there is only $1$ picture.
There are exactly $5!$ rotationally symmetric arrangements: up to permuting the colors of the balls, they have to be the arrangement represented by the picture below.
So there are $5!$ pictures representing $5!$ symmetric arrangements, which means that the remaining $\frac{15!}{3!^5} - 5!$ pictures represent asymmetric arrangements: ones that we should divide by $3$, because there are $3$ pictures of each arrangement.
So the final answer is $$ 5! + \frac13\left(\frac{15!}{3!^5} - 5!\right) = \frac{15!}{3\cdot 3!^5} + \frac23 \cdot 5! $$ (your original answer is off by $\frac23 \cdot 5! = 80$).
$\endgroup$ $\begingroup$We use the Polya Enumeration Theorem. The cycle index here is
$$Z(G) = \frac{1}{3} a_1^{15} + \frac{2}{3} a_3^5.$$
We then obtain for five colors, three each
$$[A^3 B^3 C^3 D^3 E^3] Z(G; A+B+C+D+E) \\ = [A^3 B^3 C^3 D^3 E^3] \frac{1}{3} (A+B+C+D+E)^{15} \\ + [A^3 B^3 C^3 D^3 E^3] \frac{2}{3} (A^3+B^3+C^3+D^3+E^3)^5 \\ = \frac{1}{3} \frac{15!}{3!^5} + [A B C D E] \frac{2}{3} (A+B+C+D+E)^5 \\ = \frac{1}{3} \frac{15!}{3!^5} + \frac{2}{3} 5! = 56056080.$$
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