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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

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  • $\begingroup$ Your mistake is in confusing a necklace of beads with a row of beads. With a neckace, the beads are in a circle. If the beads are all moved along the circle, so that the relative position of each bead to all of the other beads is unchanged, then the necklace is regarded as unchanged. Hence, if necklace-2 can be constructed by starting with necklace-1 and then pushing all the beads along the circle, then necklace-2 is deemed equivalent to necklace-1. $\endgroup$ Sep 3 '20 at 22:43
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Necklaces may be rotated (and sometimes mirrored, although technically that's a bracelet). So the same necklace can be made from different choices of where to place the three beads.

For instance, placing the three beads in spots 1,2 and 3 yields the same necklace as placing them in spots 4,5 and 6.

There is one necklace which can be obtained by 3 different placements, namely 1,4,7 (equal to 2,5,8 and 3,6,9), and all others may be obtained by 9 distinct placements. So of your 84, three of them are the same, and the remaining 81 go together in groups of 9. This gives 10 total possible distinct necklaces.

If there is mirroring involved too, then four of these 10 are their own mirror images, while the remaining 6 go together in pairs of two, being eachother's mirror images, resulting in 7 necklaces.

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  • $\begingroup$ thanks a lot it was indeed a good explanation to my question $\endgroup$
    – victor
    Sep 3 '20 at 5:16
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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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