<s>[INST] How many necklaces can be formed with $6$ identical diamonds and $3$ identical pearls 
Find number of ways to make a necklace (or a garland) consisting of $6$ identical diamonds and $3$ identical pearls.

I got the correct answer $7$ by taking different cases but when I applied the formula for arrangement, i.e. $$\frac{(6+3-1)!}{6!3!2}$$ I get a value which is not an integer. I divided by $6!$ and $3!$ because the diamonds and pearls are identical. Further I divided by $2$ since a necklace can be viewed from two sides.
Why am I getting a wrong answer?
 [/INST] There are  two possibilities  here, rotational symmetry  (necklace) or
dihedral  symmetry (bracelet). This is the OEIS naming convention. For the
first one  we have  the cycle index of the cyclic group:
$$Z(C_n) = \frac{1}{n} \sum_{d|n} \varphi(d) a_d^{n/d}.$$
For second one we have the cycle index of the dihedral group
$$Z(D_n) =
\frac{1}{2} Z(C_n) +
\begin{cases}
\frac{1}{2} a_1 a_2^{(n-1)/2} & n \text{ odd} \\
\frac{1}{4} \left( a_1^2 a_2^{n/2-1} + a_2^{n/2} \right)
& n \text{ even.}
\end{cases}$$
We get for $n=9$ the cycle indices
$$Z(C_9) = 1/9\,{a_{{1}}}^{9}+2/9\,{a_{{3}}}^{3}+2/3\,a_{{9}}$$
and
$$Z(D_9) =
1/18\,{a_{{1}}}^{9}+1/9\,{a_{{3}}}^{3}+1/3\,a_{{9}}
+1/2\,a_{{1}}{a_{{2}}}^{4}.$$
Then by the Polya Enumeration Theorem we require
$$[D^6 P^3] Z(C_9; D+P)$$
and
$$[D^6 P^3] Z(D_9; D+P).$$
We thus get for the first case
$$\frac{1}{9} {9\choose 6} + \frac{2}{9} {3\choose 2}
= 10$$
and the second one
$$\frac{1}{2} 10 + \frac{1}{2} [D^6 P^3] (D+P) (D^2+P^2)^4
\\ = 5
+ \frac{1}{2} [D^6 P^2] (D^2+P^2)^4
+ \frac{1}{2} [D^5 P^3] (D^2+P^2)^4
\\ = 5 + [D^3 P] (D+P)^4 = 5 + \frac{1}{2} {4\choose 3} = 7.$$
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