TITLE: How many ways to arrange $8$ read beads and $32$ blue beads into a necklace such that at least $2$ blue beads between any $2$ red beads ?
QUESTION [7 upvotes]: Zaraki wants to use $8$ indistinguishable red beads and $32$ indistinguishable blue beads to make a necklace such that there are at least $2$ blue beads between any $2$ red beads. In how many ways can he do this ?
Part of me, my solution is $30667$ and my result comes from this calculation $$\frac{\binom{23}{7}- \binom{11}{3}}{8}+ \frac{\binom{11}{3}-\binom{5}{1}}{4}+\frac{\binom 51-1}{2}+1=30667.$$
Is my solution true ?

REPLY [2 votes]: It may be of interest to  relate this problem to cycle indices and the
Polya Enumeration  Theorem (PET). Suppose we distribute  the eight red
beads on  the necklace first. Next  we distribute two  blue beads into
every space  between two red beads,  leaving $32-2\times 8  = 16$ blue
beads.  The  problem  now   becomes  equivalent  to  distributing  the
remaining  $16$ blue  beads into  the  eight slots  under rotation  or
rotation and reflection.
  To compute  the count  under rotations  we need  the  cycle index
$Z(C_8)$ of the cyclic group  on eight elements.  We now enumerate the
permutations  in  this cycle  index.   There  is  the identity,  which
contributes $a_1^8$.  A  rotation by a distance of  four maps opposite
slots to each other and creates two-cycles, giving $a_2^4.$ A rotation
by  a distance  of two  or  six creates  four-cycles, giving  $2\times
a_4^2.$  A  rotation   by  a  distance  of  $1,3,5$   or  $7$  creates
eight-cycles giving $4\times a_8.$

This gives the cycle index
$$Z(C_8) = \frac{1}{8} (a_1^8 + a_2^4 + 2 a_4^2 + 4 a_8).$$
The desired value is given by
$$[z^{16}]  Z(C_8)\left(\frac{1}{1-z}\right).$$
This is
$$\frac{1}{8} 
\left([z^{16}] \left(\frac{1}{1-z}\right)^8 + 
[z^{16}] \left(\frac{1}{1-z^2}\right)^4 + 
2 [z^{16}] \left(\frac{1}{1-z^4}\right)^2 + 
4 [z^{16}] \left(\frac{1}{1-z^8}\right)\right).$$
which yields
$$\frac{1}{8} 
\left({16+7\choose 7}
+ {8+3\choose 3}
+ 2{4+1\choose 1}
+ 4{2+0\choose 0}\right) = 30667.$$
When reflections  are included we  have dihedral symmetry and  we need
the cycle index $Z(D_8)$ of  the dihedral group on eight elements. The
additional  permutations are  four reflections  about an  axis passing
through the  midpoint of opposite  edges connecting two  slots, giving
$4\times a_2^4$ and four reflections about an axis passing through two
opposite slots giving $4\times a_1^2 a_2^3.$ Hence the cycle index is
$$Z(D_8) = \frac{1}{2} Z(C_8) +
\frac{1}{16} (4 a_2^4 + 4 a_1^2 a_2^3).$$
This yields for sixteen blue beads the count
$$\frac{1}{2} 30667
+ \frac{1}{16} 
\left(4 [z^{16}] \left(\frac{1}{1-z^2}\right)^4 + 
4 [z^{16}] \left(\frac{1}{1-z}\right)^2 
\left(\frac{1}{1-z^2}\right)^3 
\right)$$
which is
$$\frac{1}{2} 30667
+ \frac{1}{16} 
\left(4 {8+3\choose 3} +
4 \sum_{q=0}^8 {q+2\choose 2} {16-2q+1\choose´1}
\right) = 15581.$$
The sequences OEIS A032193 and 
OEIS A005514 are relevant here.

Remark. We can  also apply Burnside directly to  the cycle indices
and bypass PET. E.g. for a permutation of cycle type $a_2^4$ to fix an
assignment to the four cycles it is necessary that we
choose a number $q$ of blue beads for each the four two-cycles
and then place  twice that number $q$ of beads  on the four two-cycles
with $q$  beads in each  of the  two slots on  a two cycle.  
Choosing a pair has generating function
$$\frac{1}{1-z^2}$$
and the total contribution is
$$[z^{16}]
\frac{1}{1-z^2}
\frac{1}{1-z^2}
\frac{1}{1-z^2}
\frac{1}{1-z^2}.$$
Similarly an assignment  for a permutation of cycle  type $a_4^2$ that
is fixed by  this shape of permutation means that we  have to choose a
number $q$  of blue beads  for each of  the two cycles and  then place
four times  that number on  each cycle with  $q$ beads in each  of the
slots on the four-cycle. This gives
$$[z^{16}]
\frac{1}{1-z^4}
\frac{1}{1-z^4}.$$
This can be continued and it is essentially the mechanism by which PET is proved starting from Burnside.

There are many more related links at 
MSE Meta on Burnside/Polya.