Rattles with beads or necklace with beads? I came across this problem in a book called limits, sequences combinations great book for intro to combinatorics .

A rattle consists of a ring with $3$ white beads and $7$ red ones strung on it. Some rattles seemingly different can be made identical by arranging the rings and moving the beads in a suitable manner (rotation or flipping). Find the number of different rattles .

I of course thought polya enumeration on this one but was thinking it can be done case by case without being too messy . Can anyone help? Also this is essentially  the same problem as a necklace with $n$ beads, $k$ colors is it not ? 
 A: A hand count is not hard.  You have to find a way to organize it so you count each configuration only once.  We get one configuration for each weak partition of $7$ into $3$ parts.  We get
$$7,0,0\\6,1,0\\5,2,0\\5,1,1\\4,3,0\\4,2,1\\3,3,1\\3,2,2$$ for eight possibilities.  Clearly these are all distinct.  You need to convince yourself that there are not two configurations for a partition, but rotation and flipping give six configurations, which matches the number of orders of a partition.
A: Let me just observe that with $(3,10) = 1$ and $(7,10) = 1$ Polya will
be  very simple  in  this  case. The  OEIS  uses  necklace for  cyclic
symmetry and  bracelet for dihedral, so  we have a bracelet  here. The
cycle index is from first principles
$$Z(D_{10}) = \frac{1}{20} \sum_{d|10} \varphi(d) a_d^{10/d}
+ \frac{1}{4} a_2^5 + \frac{1}{4} a_1^2 a_2^4.$$
With the two colors our answer is given by
$$[W^3 R^7] \frac{1}{20} \sum_{d|10} \varphi(d) (W^d+R^d)^{10/d}
\\ + [W^3 R^7] \frac{1}{4} (W^2+R^2)^5
\\ + [W^3 R^7] \frac{1}{4} (W+R)^2 (W^2+R^2)^4.$$
From  the first  term  only  $d=1$ contributes,  the  second does  not
contribute at  all and from the  third the first term  must contribute
$WR.$ We get
$$[W^3 R^7] \frac{1}{20} (W+R)^{10}
+ [W^2 R^6] \frac{1}{2} (W^2+R^2)^4
\\ = [W^3 R^7] \frac{1}{20} (W+R)^{10}
+ [W^1 R^3] \frac{1}{2} (W+R)^4.$$
The desired count is thus
$$\frac{1}{20} {10\choose 3} + \frac{1}{2} {4\choose 1}
= 8.$$
