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Find the number of ways to simultaneously select $3$ pebbles from $9$ pebbles that are arranged in a circle such that no two of the three selected pebbles are adjacent to each other, and one selection is the same as another rotation if the former can be rotated to form the latter.

My Solution: At first, we can pick any $9$ points. Then, any $6$ points. And at last, $4$ points. This turns out be $6\cdot4\cdot9$. However, we have over-counted. We divide this by $3$! because we do not care about the order. Lastly, we divide by $9$ to account for the last part of the problem. Answer: $4$

Is this correct? And is there an easier way using combinatorics?

Kenta S
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Akshat Mundra
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    I'm not sure where you get the "Then, any 6 points" (which I assume means "any of 6 points") -- I assume it means "not the picked point or either of its two neighbors", and that makes sense. But the "and at last, 4 points" bothers me. If you've picked 1 and 3, 5,6,7,8 remain -- i.e., four choices. But what if you've picked 1 and 4. What choices remain for the third point? I only see 6, 7, 8. – John Hughes Nov 11 '19 at 18:25
  • "and at last 4 points" ... But if second point is chosen far from the "border" with the first, then for the third point you are left with 3 choices (if I can see that well). – G Cab Nov 11 '19 at 18:26
  • This question has not been asked before. The $6$ non selected pebbles are partitioned into $3$ positive parts. There are three such partitions, namely $(4,1,1)$, $(3,2,1)$, $(2,2,2)$. The first and the third give one of your configurations, the second gives two (mirror symmetric ones). Your answer $4$ therefore is correct. – Christian Blatter Nov 11 '19 at 19:12
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    Not sure that the linked duplicate is actually a duplicate as that question and all its answers do not consider rotated configurations to be equivalent. In the linked question you can easily handle rotational equivalence by dividing the given answers by 20, but the analogous method (dividing by 9) won't work here because not all orbits under rotation have size 9 (the configurations that look like equilateral triangles, in particular). – Will Orrick Nov 11 '19 at 19:13
  • An alternative to Christian Blatter's answer: there are $\binom{9}{3}-9\cdot5-9=30$ configurations with none adjacent. Three of these look like equilateral triangles, and are equivalent under rotation. The other 27 are overcounted by a factor of 9 due to rotations, giving $1+27/9=4$ as the answer. Although the OPs answer is the right number, I agree with the comments of John Hughes and G Cab that the method is suspect. – Will Orrick Nov 11 '19 at 19:17
  • One can also account for the symmetry using Burnside's lemma: of the nine elements of the rotation group, the identity fixes all 30 configurations and clockwise rotation by 3 or 6 spaces fixes the three equilateral triangle configurations. Hence the number of orbits is $\frac{1}{9}(30+2\cdot3)=4$. – Will Orrick Nov 11 '19 at 19:50

2 Answers2

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We may also use the Polya Enumeration Theorem. Selecting three non-adjacent pebbles leaves a total of six non-selected ones that must be placed in three available slots with the cyclic group $C_3$ acting on them and at least one pebble in every slot. The cycle index is

$$Z(C_3) = \frac{1}{3} a_1^3 + \frac{2}{3} a_3.$$

We then have

$$[z^6] Z\left(C_3; \frac{z}{1-z}\right) = [z^6] \left(\frac{1}{3} \frac{z^3}{(1-z)^3} + \frac{2}{3} \frac{z^3}{1-z^3} \right) \\ = \frac{1}{3} [z^3] \frac{1}{(1-z)^3} + \frac{2}{3} [z^3] \frac{1}{1-z^3} = \frac{1}{3} {3+2\choose 2} + \frac{2}{3} [z^1] \frac{1}{1-z} \\ = \frac{1}{3} \times 10 + \frac{2}{3} = 4.$$

Marko Riedel
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You have obtained the correct answer, but by accident. As pointed out in the comments by John Hughes and G Cab, it is not the case that after picking the first two points you always have four choices for the third point: in some cases you have only three choices. This reduces the number of configurations that you have prior to accounting for equivalences. On the other hand, the size of the equivalence classes is not always $54=6\times9$: for configurations that look like an equilateral triangle, there are only three rotated versions of the configuration, not nine. It seems that by lucky coincidence, these two overestimates cancel each other out. I don't see any deep reason for this, and you can check that your method doesn't work when the number of pebbles is changed from 9 to 8 or 10 (or to 6 or 12, which are similar to 9 in that they also have equilateral triangle configurations).

A method that does work is to use the methods given here or here to compute the number of ways of selecting three pebbles, no two adjacent. Using Brian Scott's method, you get $$ \binom{9}{3}-9\cdot(9-4)-9=30. $$ Now the three configurations that look like equilateral triangles are equivalent under rotation and the 27 remaining configurations come in equivalence classes of size 9. This gives $1+\frac{27}{9}=4$ equivalence classes.

You can also account for the symmetry using Burnside's lemma: of the nine elements of the rotation group, the identity fixes all $30$ configurations and clockwise rotation by $3$ or $6$ spaces fixes the three equilateral triangle configurations. Hence the number of orbits is $$ \frac{1}{9}(30+2⋅3)=4. $$

Will Orrick
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