Possible combinations of $4$ pieces of $A$ and $4$ pieces of $B$ I've worked out the question by trial and error, but I'm just wondering whether there is a more formal method of solving this problem. The problem is as follows:
$4$ pieces of white cubes and four pieces of black cubes are equally spaced around a chain. There is no clasp, so the pieces may be rotated. Also the chain can be turned over. In how many ways can the $8$ cubes be placed around the chain?
I found $8$ solutions: $BBBBWWWW, BBBWWWBW, BBBWWBWW, BBWWBBWW, BBWBBWWW, BBWWBWBW, BWBWBWBW, BBWBWWBW$
where $B$ denotes black cubes and $W$ denotes white cubes.
Any thoughts on how to answer this question more methodically?
 A: As per Burnside's counting theorem, here is how you could do this. It may seem long as I am explaining every step. But once you start doing, it becomes much simpler.
As there are $8$ cubes, you can see it as a regular octagon with $8$ vertices - $4$ to be colored in black and rest $4$ in white.
Rotational symmetry in a regular octagon - any rotation by $45^0$ gives you no change in the appearance of the octagon. So there are in total $8$ rotational symmetry.
Now let's look at reflection. There are $8$ axes of symmetry in total. $4$ of them are axes joining the two opposite vertices and $4$ of them are axes joining the midpoints of two opposite sides.
That is why a regular octagon is called Dihedral group $D8$ of order $16$ ([wikipedia][1]
Now we will start with counting. As per Burnside Lemma, you have to count all group actions that keep the colors fixed by rotation and by reflection.
A) Rotation
i) In original position - you can color $4$ vertices each with black and white in $\dfrac{8!}{2!2!} = 70$ ways.
ii) Rotation by $45^0$ - no arrangement in $(i)$ gives you same fixed colors when you rotate. So will be true for $135^0, 225^0$ and $315^0$,
iii) Rotation by $90^0$ - You can only get fixed colors when vertices of alternate colors are same. This can be done in $2$ ways. Same will be true for $270^0$. So, $= 4$ ways.
iv) Rotation by $180^0$ - You can get fixed colors when opposite vertices are of the same color. So you have $2$ pairs each of black and white. This can be done in $\dfrac{4!}{2!}{2!} = 6$ ways.
By rotation, number of group actions that keep the colors fixed $= 80$
B) Reflection
i) Rotation about axis through opposite vertices - each pair of vertices on $3$ perpendicular lines to the axes will have to have the same color within the pair for colors to be fixed during reflection along the axis which also means the $2$ vertices on the axes will have the same color too. So you again have $2$ pairs each of black and white. This can be done in $\dfrac{4!}{2!}{2!} = 6$ ways as we saw earlier.
But as we have $4$ axes, number of ways $= 4 \times 6 = 24$.
ii) Rotation about axis through midpoints of opposite sides - each pair of vertices on $4$ perpendicular lines to the axes will have to have the same color within the pair for colors to be fixed during reflection along the axis. With $4$ such axes, again, number of ways $= 24$.
So by reflection, you get $24+24 = 48$ group actions that keep the colors fixed.
Adding both $(A)$ and $(B)$, you get $80 + 48 = 128$ group actions.
Now as per Burnside Lemma,
Number of distinct arrangements = (total number of group actions to keep the colors fixed / order of the group)
So number of distinct arrangements = $\dfrac{128}{16} = 8$
[1]: https://en.wikipedia.org/wiki/Octagon#:~:text=In%20geometry%2C%20an%20octagon%20(from,alternates%20two%20types%20of%20edges.
