There are 8 red beads and 32 blue beads. How many necklaces (which can be rotated) are there such that between any two red beads there are at least 2 blue beads.
I reduced the problem to finding the set of solutions of
$$x_1+ x_2 +\cdots +x_{8} = 16$$
Where no 8-tuple is a cyclic permutation of the other. But I really can't see how to make any further progress.
Here’s a solution that doesn’t apply Burnside’s Lemma (though the idea of that lemma is hidden in it).
Suppose you found all the solutions to $x_1+ x_2 +\cdots +x_{8} = 16$, wrote them in a list, and drew the corresponding necklace for each one. There would be $16+8-1\choose8-1$ solutions and pictures of necklaces.
Many necklaces would appear eight times in your list. If every necklace appeared eight times, the number of different necklaces would be easy to calculate; it would be $\frac{1}{8}\cdot{16+8-1\choose8-1}$.
Unfortunately, not every necklace appears eight times. Some necklaces appear only four times, some twice, and one necklace would appear exactly once. Here’s a picture of just a few solutions and (straightened-out) necklaces, grouped by matching necklaces.
So here’s an idea: carefully duplicate some of the solutions just as necessary so that you have a list where all the necklaces do appear exactly eight times.
If a necklace appears only four times, list each solution with that necklace twice. If a necklace appears just twice, list each solution for that necklace four times, and include the solution that gives one necklace ($x_i=2$) eight times instead of once.
You’ll end up with more than $16+8-1\choose8-1$ “solutions.” How many more - how can you count the number of “solutions-plus-repeats”?
The solutions that gave necklaces that appeared only four times are solutions of the form $x_1+ x_2 +x_3+x_4+x_1+x_2+x_3+x_4 = 16$, that is, they are solutions of $x_1+ x_2 +x_3+x_4=8$. We know there are $8+4-1\choose4-1$ of those. These solutions need to be listed twice, not once, so add $8+4-1\choose4-1$ to your tally.
But oops, some of these solutions actually gave necklaces that appeared just twice, and we ultimately need those to be listed four times (not twice, which they are so far) so we need to add an additional two copies of each solution to $x_1+x_2=4$. That’s another $2\cdot{4+2-1\choose2-1}$ to the tally. And finally, add four more copies of the solution $x_i=2$ so we have eight of those.
The final tally of solutions-plus-duplicates is
$$ \begin{align}&{16+8-1\choose8-1}+{8+4-1\choose4-1}+2\cdot{4+2-1\choose2-1}+4\\=&{23\choose7}+{10\choose3}+2\cdot{5\choose1}+4\\=&245,336. \end{align}$$
The number of necklaces is one-eighth of this, or $30,667$.
I think your problem reduction is right.
Without taking rotation groups into account, I would say that you need to place 16 blue beads in 8 groups.
Number of ways of distributing n identical objects among r groups
But this does not take into account the rotation symmetry (don't forget the mirror symmetry if rotation is allowed).
Obviously this is not a high school question in that case.
Would you know necklace with 8 white and 16 blue beads?
A blue red blue sequence would be replaced by a white bead.
