Circular arrangements of identical objects Q> In how many ways can 5 identical red beads, 3 identical green beads and 2 identical blue beads be arranged in a necklace? 
 A: Since this question is about counting orbits under a finite group, it is useful to apply Burnside's lemma, which says that the number of orbits under a finite group acting on a finite set is always equal to the average, taken over the elements $g$ of the group, of fixed points under $g$. The group here is the dihedral group $D_{10}$ of order$~20$, acting on circular arrangements of $10$ elements chosen from $3~$colours.
The case $e=g$ always contributes, so start counting the number of arrangement disregarding symmetry. The number then is the trinomial coefficient
$$
  \binom{10}{2,3,5}=\frac{10\times9}{2\times1}\times\frac{8\times7\times6}{3\times2\times1} = 2520
$$
The other rotations in $D_{10}$ have order $2,5$ or$~10$, but since nonre of these divide all the required coloiur frequencies (i.e., $2,3$ and$~5$, which indeed have $\gcd(2,3,5)=1$), so they don't contribute any fixed colourings. There remain the reflections, which come in two conjugacy classes: those that fix $2$ of the $10$ beads, and those that fixed none (each exchanging $5$ pairs of beads). The latter don't have any fixed colouring of the required type either (the odd-frequency colours cannot be used just for a subset of the interchanged pairs). So let $g$ be a reflection that fixes two (opposite) beads, and partitions the remainder into $4$ pairs. To get an invariant colouring, both fixed beads must get a different odd-frequency colour (red or green), and then the $4$ pairs must be colours as $1$ pair blue, $1$ pair green, and $2$ pairs read. The number of such colourings is $2\times\binom4{1,1,2}=24$ (the initial factor $2$ is from the colouring of the fixed beads of $g$). 
Now to wrap things up, compute the average over the group, given that there is one identity, and $5$ group elements of the contributing tpye $g$ above:
$$
  \frac{1\times2520+5\times24}{20}=132
$$
which fortunately is integer.
A: If you consider placing $10$ beads in a straight line with order, then the number of ways will be
$${10 \choose 5} \cdot {5 \choose 3} = 2520$$
Beads with same color are identical, so you should use the "choose" operation. (You can do this computation in any order, like choosing green ones first, blue ones next, they will all yield same result.)
Since you can rotate a necklace, the final number arrangement ways are $252$.

Thanks @Alexey Burdin pointing out. This case is lucky because sometimes when we rotate for $n$ times ($n$ is the number of total beads), we have multiple times to get back to the original way. To get a beginning, we look at OP's case and only consider red ones. Do some attempt on paper, we can find out that this occur iff each of red ones are separated by exactly one of other color. Then each even number of rotation we have a problem.
But the green ones don't have this problem. Let's label the position in straight line of green ones as $a,b,c$, and "rotate" (meaning doing modulo asthmatics) $n$ times. We get $a+n\equiv x, b+n\equiv y, c+n\equiv z$, where $x,y,z$ are exactly $a,b,c$ in any order. Therefore, $3n \equiv 0 \mod 10 \Rightarrow n \equiv 0 \mod 10$.
In short, if at least one color has number that not divide total number, the case is easy. In general, we have to figure out some specific patterns that are duplicated for multiple times.
