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We have a set $S$ with $E$ elements of which only $N$ are unique. We of course know how many repetitions of each of the $N$ elements are present: element $s_i$ is repeating $t_i$ times.

I would like to count the number of ways we can divide the $E$ elements in $N$ blocks of size $k_1, k_2, \cdots , k_N$ when the elements within the block are indistinguishable.

If the $E$ elements are all unique, we can answer directly using the Bell Polynomials.

Do you think is it possible to extend the above result ?

Bilbottom
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Ninja Warrior
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  • Just to be sure I understand if we have the multiset ${2\bullet3, 3\bullet4}$ meaning $2$ copies of $3$ and $3$ copies of $4$, and $N=3,$ then $2\bullet3,2\bullet4,1\bullet4$ is an acceptable partition, correct? Does order matter? Is the previous partition the same as $2\bullet3,1\bullet4,2\bullet4?$ – saulspatz Aug 19 '18 at 16:15
  • In your example $k_1=2,k_2=2,k_3=1$ and $S={3,3,4,4,4}$ the possible partitions are 5: $(3,3)(4,4)(4)$ ; $(3,4)(3,4)(4)$ ; $(3,4)(4,4)(3)$ ; $(4,4)(3,4)(3)$ ; $(4,4)(3,3)(4)$ – Ninja Warrior Aug 19 '18 at 17:07

1 Answers1

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It appears that we have a simplified version of the computation from the following MSE link. Using the notation that was presented there we obtain the closed form

$$\left[\prod_{k=1}^l A_k^{\tau_{k}}\right] \prod_{k=1}^m Z\left(S_k; \sum_{k'=1}^l A_{k'}\right)^{\sigma_k}.$$

In terms of combinatorial classes we have made use of the unlabeled class

$$\def\textsc#1{\dosc#1\csod} \def\dosc#1#2\csod{{\rm #1{\small #2}}} \textsc{SEQ}_{=\sigma_k} \left(\textsc{MSET}_{=k} \left(\sum_{k'=1}^l \mathcal{A}_{k'}\right)\right).$$

Note that the cycle index will create the intermediate multisets during evaluation.

Marko Riedel
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