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$n$ distinct objects have $n!$ (linear) permutations and thus $(n-1)!$ circular permutations.

Now consider $n$ objects, some identical, $r_1$ of the first kind, $r_2$ of the second kind, ..., $r_k$ of the $k$th kind. These $n$ objects have $\frac{n!}{r_1!r_2!\dots r_k!}$ (linear) permutations.

Can we likewise reason that these $m$ objects have $\frac{(n-1)!}{r_1!r_2!\dots r_k!}$ circular permutations? I think the answer is no, but can someone explain the intuition why the reasoning that worked earlier doesn't work here? Also, what is the correct number of circular permutations for these $m!$ objects?

(I am hoping for an answer that's suitable for high school students. Thanks.)

Amir
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    The correct calculation will allow for the fact that under certain circimstances, some arrangements of the circle will be indistinct from each other because for TWO reasons: Rotational symmetry of the circle, and rotational symmetry of the pattern $m_1,m_2,m_4...$ due to indistinct elements. Therefore the prior method will have subtracted these twice so you will need to add them back in, by the inclusion-exclusion principle. – it's a hire car baby Jul 16 '16 at 12:07
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    Hint: try some small examples. The case $k = 1$, $m > 1$ should give $1$ permutation, but your formula gives $1/m$ (not even a whole number). A better guess based on the argument for the case with no identifications would be $(m-1)!/((r_1-1)!\ldots r_k!)$, but that will be an overestimate unless $r_1 = 1$: try the case $k = 2$, $r_1 = 2$, $r_2 = 3$ to see why. – Rob Arthan Jul 16 '16 at 12:13
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    This can be done using Burnside's Lemma, as you are counting equivalence classes of permutations of elements in a circle under rotational equivalence. See https://en.m.wikipedia.org/wiki/Burnside%27s_lemma – Tad Jul 16 '16 at 18:47
  • This problem recently appeared at this MSE link. – Marko Riedel Jul 16 '16 at 19:38
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    This question has been answered here: https://mathoverflow.net/questions/86709/circular-permutations-with-identical-objects – Alexander Oct 03 '20 at 00:30

1 Answers1

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Hope the answer below, provided after 7 years, is useful for readers who are less familiar with advanced counting techniques.

The closed-form answer is

$$\frac{1}{n}\,\sum_{d\mid\gcd(r_1,r_2,\ldots,r_k)}\,\phi(d)\,\binom{n/d}{r_1/d,r_2/d,\ldots,r_k/d}$$

where $\phi(d)$ is the Euler's totient function (returning the number of positive integers $m$ less than $n$ that are coprime to $n$, i.e. $\gcd(n,m)=1$; e.g. $\phi(6)=2, \phi(7)=1, \phi(8)=3$).

For the case of $\gcd(r_1,r_2,\ldots,r_k)=1$, the answer is simplified to

$$\frac{1}{n}\binom{n}{r_1,r_2,\ldots,r_k}.$$

Note that this is $n\text{th}$ times smaller than the number of linear permutations with repetitions.

In particular, for $r_1=r_2=\ldots=r_k=1$, one gets the number of circular permutations of $n$ distinct objects:

$$\frac{1}{n}\binom{n}{1,1,\ldots,1}=(n-1)!.$$

For example, for $n=6, r_1=r_2=r_3=2$, the answer is

$$\frac{1}{6} \left(\phi(1)\binom{6/1}{2/1,2/1,2/1}+\phi(2)\binom{6/2}{2/2,2/2,2/2} \right) =\frac{1}{6} \left (1\times \frac{6!}{2!2!2!} +1\times \frac{3!}{1!1!1!} \right)=16.$$

The above closed-form solution can be obtained by applying the Pólya’s Inventory Theorem to the cycle index polynomial for the cyclic group of order $n$, which is given by

$$\frac{1}{n}\,\sum_{d\mid n}\,\phi(d) x^\frac{n}{d}_d.$$

By replacing each $x_d$ by $\sum_{i=1}^{k}\, z^d_i$, the $d$th power sum of the weights $z_1, z_2, \ldots, z_k$, in the above polynomial, one gets

$$\frac{1}{n}\,\sum_{d\mid n}\,\phi(d) \left(\sum_{i=1}^{k}\, z^d_i \right)^\frac{n}{d}.$$

The Pólya’s Inventory Theorem states that the coefficient of the monomial $z_1^{r_1}z_2^{r_2}\ldots z_k^{r_k}$ in the above nested polynomial is the number of circular permutations with repetitions.

Again consider our example above, $n=6, r_1=r_2=r_3=2$. The nested polynomial is obtained as follows:

$$\frac{1}{6} \left(\phi(1)\left(z^1_1+z^1_2+z^1_3\right)^\frac{6}{1}+\phi(2)\left(z^2_1+z^2_2+z^2_3\right)^\frac{6}{2}+\phi(3)\left(z^3_1+z^3_2+z^3_3\right)^\frac{6}{3}+\phi(6)\left(z^6_1+z^6_2+z^6_3\right)^\frac{6}{6}\right)$$

$$=\frac{1}{6} \left( \left(z_1+z_2+z_3\right)^6+\left(z^2_1+z^2_2+z^2_3\right)^3+2\left(z^3_1+z^3_2+z^3_3\right)^2+2\left(z^6_1+z^6_2+z^6_3\right)\right)$$

The monomial $z^2_1z^2_2z^2_3$ only appears in the first and second terms with the coefficients $\binom{6}{2,2,2}$ and $\binom{3}{1,1,1}$, respectively.

Amir
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