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How many different necklaces can be made from beads. If we have

  • 10 green beads
  • 5 red beads
  • 3 yellow beads

And we will use all the beads for creation of necklace.

Disclaimer: Not a homework question, just preparing for test

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    $\begingroup$ $\frac{18!}{10!.5!.3!}$ $\endgroup$ Jan 9 '18 at 11:38
  • $\begingroup$ ${P(10,5,3)}$ is when we are not counting the axis symmetry? $\endgroup$ Jan 9 '18 at 11:45
  • $\begingroup$ Then it is a restriction that you have to mention. $\endgroup$ Jan 9 '18 at 11:47
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There are $\frac{18!}{10!\times 5!\times 3!}$ different ways to arrange the beads in a row.

To see this, start by labelling the beads to make them all different, and now there are $18!$ ways to order the different beads. However, every arrangement of unlabelled beads corresponds to $10!\times 5!\times 3!$ arrangements of the labelled beads (you can rearrange the green labels in $10!$ ways, etc.).

But we want the number of necklaces, not the number of ways to arrange the beads in a row. Now each necklace corresponds to $18$ ways of arranging the beads in a row - break the necklace at any point and start the row there. These $18$ arrangements are always all different (this is not necessarily true for a general necklace problem, but it is here). This is because if two different places, $k$ apart, to break the necklace gave the same arrangement, there would always be another red bead $k$ places after every red bead, and since $5$ and $18$ are coprime this is not possible.

Thus we have to divide the number of arrangements by $18$ to get the number of necklaces, which is therefore $\frac{17!}{10!\times 5!\times 3!}$

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  • $\begingroup$ When I imagine this necklace, I imagine the beads all fall to the bottom in front of the neck, so that they're in a line rather than a loop. It depends what the OP meant, of course. $\endgroup$
    – Myridium
    Jan 9 '18 at 12:17
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    $\begingroup$ @Myridium "necklace" has a standard meaning in this sort of problem: en.wikipedia.org/wiki/Necklace_(combinatorics) $\endgroup$ Jan 9 '18 at 12:21
  • $\begingroup$ Huh, fair enough. Thanks, I wasn't aware. $\endgroup$
    – Myridium
    Jan 9 '18 at 12:21
  • $\begingroup$ In general, couldn't we have a necklace, say $RBRBRBRB$ where there are fewer than $18$ distinct rotational variants? I'm not sure how to account for this, so I've deleted my answer for now. $\endgroup$
    – Myridium
    Jan 9 '18 at 12:34
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The answer is $$\frac{18!}{10!.5!.3!} =2450448$$ Here is how we count the necklaces.

For the first bead you have $18$ choices, for the second one you have $17$ choices, and so forth ...until the last bead for which you have only $1$ choice.

That makes it $18!$ if all the beads were different. Since we can rearrange the $ 10$ green beads in $10!$ ways and rearrange the $5$ red beads in $5!$ ways and the $3$ yellow beads in $3!$ ways, we divide the $18!$ by $(10!5!3!)$ and get the result.

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  • $\begingroup$ $\frac{18!}{10! \cdot 5! \cdot 3!} = 2450448$ if I'm not doing something stupid. $\endgroup$
    – Myridium
    Jan 9 '18 at 12:14
  • $\begingroup$ You are correct. I fixed my mistake. Apparently my calculator is still sleeping. $\endgroup$ Jan 9 '18 at 12:25
  • $\begingroup$ Your answer and that of @EspeciallyLime contradict each other. Do you find that conspicuous? $\endgroup$ Jan 9 '18 at 13:41
  • $\begingroup$ He is assuming that beads can move freely around the necklace which may or may not be true. Normally there is a pendent which makes the free movement impossible. $\endgroup$ Jan 9 '18 at 13:54

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