<s>[INST] A necklace is made up of 3 beads of one color and 6 beads of another color A necklace is made up of 3 beads of one color and 6 beads of another color beads of same color are identical the number of necklace that are possible ?
I have attempted the question in this way :
there are total 9 places and in which 3 objects of one kind and 6 objects of other kind are repeating so number of ways would be $\frac{9!}{3!.6!}$ = 84  which turns out to be a wrong answer when i inspected so please help me where I am wrong.
answer given is 7 possible necklaces
 [/INST] Consulting the following fact sheet on necklaces and
bracelets
we get for the cycle index of the cyclic group
$$Z(C_n) = \frac{1}{n} \sum_{d|n} \varphi(d) a_d^{n/d}.$$
We will apply PET so we need
$$Z(C_9) = \frac{1}{9} (a_1^9 + 2 a_3^3 + 6 a_9).$$
Our answer is then given by
$$[X^3 Y^6] Z(C_9)(X+Y) \\ =
\frac{1}{9} [X^3 Y^6] (X+Y)^9
+ \frac{2}{9} [X^3 Y^6] (X^3+Y^3)^3
+ \frac{2}{3} [X^3 Y^6] (X^9+Y^9).$$
This yields
$$\frac{1}{9} {9\choose 3} + \frac{2}{9} [X Y^2] (X+Y)^3
= \frac{28}{3} + \frac{2}{9} {3\choose 1} =
\bbox[5px,border:2px solid #00A000]{10.}$$
With reflections allowed we get bracelets and the cycle index
$$Z(D_9) = \frac{1}{2} Z(C_9) + \frac{1}{2} a_1 a_2^4$$
which now yields
$$\frac{1}{2} 10 + \frac{1}{2} [X^3 Y^6] (X+Y) (X^2+Y^2)^4
\\ = 5 + \frac{1}{2} [X^3 Y^6] X (X^2 + Y^2)^4
\\ = 5 + \frac{1}{2} [X^2 Y^6] (X^2+Y^2)^4
= 5 + \frac{1}{2} [X Y^3] (X+Y)^4
= 5 + \frac{1}{2} {4\choose 1} =
\bbox[5px,border:2px solid #00A000]{7.}$$
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