Suppose you have beads with colors and numbers on them. There are 8 colored white, 6 colored black, and 3 colored red. The white are numbered 1 to 8, the black numbered 1 to 6, and the red numbered 1 to 3. They are then arranged in a circle on the floor, such that no two of the same color are adjacent. How many arrangements are possible, if rotations of the beads are indistinguishable?
I tried the following calculation which is clearly false since it results in a non-integer:
First we count with position indices. If we place the white beads first at positions 1, 4, 6, 8, and so on, there are 8! ways to fill these positions. If position 2 is black, then position 3 is red, so there are 3 ways to fill position 3, and then 8! ways to fill the remaining 8 positions arbitrarily. We may make position 2 red and 3 black.
So I count $8!8!\cdot 2 \cdot 3$ ways to arrange this with indices. Since there are 17 positions, then every rotation is equivalent, and we get $8!8!\cdot 2\cdot 3/17$.
But I can't spot a logical mistake here, and I don't see another way to count the arrangements.
Oh I just spotted that the second counting of 8! is invalid since one may pick any color (not just white) for position 2. But even correcting for this we then would say there are $8!\cdot 3\cdot 6\cdot 2 \cdot 7! / 17$ arrangements, still not an integer.
Best Answer
I think I realized the answer, which is that what I counted actually doesn't need to be divided by the number of rotations. Rather, it already makes one particular rotation canonical by placing the two non-white beads in positions 2 and 3.
So in fact $8!\cdot 3\cdot 6\cdot 2\cdot 7!$ is the solution.