Understanding Burnside's theorem and using it to solve an elementary problem. I would like to know how to solve this problem using Burnside's theorem:
A bracelet is to be made by threading four identical red beads and four identical yellow beads onto a hoop. How many different bracelets can be made?
Furthermore how do I know when to apply this theorem? How exactly does it work? 
 A: Before we start let us observe that OP asks for a bracelet rather than
a  necklace which  means we  have dihedral  symmetry rather  than just
rotational symmetry. We can use either the Burnside lemma or the Polya
Enumeration Theorem on this, both  require the cycle index $Z(D_8)$ of
the dihedral group $D_8$. 
Now to compute the cycle index we start with the rotations. Let the
slots  be  numbered  zero to  seven.   There  is  the identity  for  a
contribution of  $a_1^8.$ The rotation that  takes $0$ to  $1$ makes a
contribution  of  $a_8.$  The  one  that  takes $0$  to  $2$  makes  a
contribution of  $a_4^2.$ The  one that takes  $0$ to  $3$ contributes
$a_8.$ The one that takes $0$  to $4$ contributes $a_2^4$. From $0$ to
$5$ we get $a_8.$  From $0$ to $6$ we get $a_4^2.$  Finally $0$ to $7$
contributes $a_8.$
Next do  the reflections.  We have four  reflections about  an axis
passing  through  the  centers  of  two opposite  edges,  each  giving
$a_2^4.$ We also have four  reflections passing through the centers of
opposite vertices for a contribution of $a_1^2 a_2^3.$
  Collect the  reflections and  the rotations  to obtain  the cycle
index
$$Z(D_8) = \frac{1}{16}
(a_1^8 + a_2^4 + 2 a_4^2 + 4 a_8 + 
4 a_2^4 + 4 a_1^2 a_2^3)
\\ = \frac{1}{16}
(a_1^8 + 5 a_2^4 + 2 a_4^2 + 4 a_8 + 
4 a_1^2 a_2^3).$$
Now  to  apply  Burnside  we  must compute  the  number  of  colorings
containing four red beads and four  yellow beads fixed by each type of
permutation. For  $a_1^8$ we must  choose four red  singletons, giving
${8\choose 4}.$ For  $a_2^4$ we must choose two of  the four cycles to
be red, for  a contribution of $5\times {4\choose  2}.$ For $a_4^2$ we
have two possiblities giving  $2\times 2.$ No contribution from $a_8.$
Finally for $a_1^2  a_2^3$ the singletons are either  both red or both
yellow, giving $2\times 4\times {3\choose 2}.$ We get for the answer
$$\frac{1}{16} 
\left({8\choose 4} + 5\times {4\choose 2}
+ 2\times 2 + 2\times 4\times {3\choose 2}\right)
= 8.$$
Here we used the fact that to be fixed by a given permutation an assignment must be constant on the cycles.
The same  result can  be obtained from  the Polya  Enumeration Theorem
(PET). We have
$$Z(D_8)(R+Y) = 
1/16\, \left( R+Y \right) ^{8}+{\frac {5\, \left( {R}^{2}+{Y}^{
2} \right) ^{4}}{16}}\\+1/8\, \left( {R}^{4}+{Y}^{4} \right) ^{2}
+1/4\,{R}^{8}+1/4\,{Y}^{8}+1/4\, \left( R+Y \right) ^{2}
 \left( {R}^{2}+{Y}^{2} \right) ^{3}
\\= {R}^{8}+{R}^{7}Y+4\,{R}^{6}{Y}^{2}+5\,{R}^{5}{Y}^{3}+8\,{R}^{4}
{Y}^{4}\\+5\,{R}^{3}{Y}^{5}+4\,{R}^{2}{Y}^{6}+R{Y}^{7}+{Y}^{8}.$$
We see that $$[R^4 Y^4] Z(D_8)(R+Y) = 8.$$
Here we have used the standard substitution $a_d = R^d + Y^d.$
Remark. When manual computation becomes too cumbersome there is
the formula
$$Z(C_n) = \frac{1}{n}\sum_{d|n} \varphi(d) a_d^{n/d}$$
for the  cycle index of the  cyclic group $Z(C_n)$ of  order $n$ which
leaves the reflections, which are easy.
