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I want to decompose given list $P=\{1,2,3,...,n\}$ into three subsets $Q,R,S$ considering without ordering of a given set.

Let me explain my situation by counting the rank $3$ case.

Let $P=\{1,2,3\}$ I want to divide $P=QRS$ with $Q,R,S$ non-empty sets without considering ordering. Simply in this case there are $3!=6$ ways with $(Q,R,S) = \{(1,2,3), (1,3,2), \cdots \}$.

Let $P=\{1,2,3,4\}$. Then at least one element of its subset contains two elements. Let me choose my $Q$ with choosing two elements $1,2$. Then since I am not considering the order I just denote my $Q$ as $Q=12$. Considering the whole set $P=(Q,R,S)$. There are only two cases {(12,3,4), (12,4,3)}. So consider the whole partition decomposition, I have $3 \times \binom{4}{2} \times 2 = 36$

For rank $5$ case there are $2+2+1, 3+1+1$ case and for each case the number of partitions are $\binom{5}{2} \binom{3}{2} \times 3 + \binom{5}{3} \times 2 \times 3 = 90+60=150$.

I am somehow exhausted to compute all the higher rank cases, And there seems a closed general formula for the general case. Is there any closed formula for the general $n$ case?

phy_math
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  • @phy_math You're still considering orderings of the three distinct sets. You should write, you're not considering "ordering of elements within any set". – MyMolecules Dec 23 '21 at 12:54

1 Answers1

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You are looking at $3!\cdot {n\brace 3}$ where ${n \brace k}$ is the Stirling number of the second kind. You can write this as $$3^{n}-3\cdot 2^{n}+3.$$

Phicar
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  • Hello, @Phicar, Is there any reference for deriving the above formula? ; I want to extend this to four decompositions etc and so on. It seems that If I understand this particular general formula for three decomposition, I might can do for next 4 decomposition, 5 decomposition and maybe get general. – phy_math Dec 24 '21 at 11:42
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    @phy_math Yes, you can show this yourself using inclusion-exclusion. Say you want $k$ blocks. Place the $n$ elements in a line and one by one assign a number from $1$ to $k$ in $k^n$ ways. Notice you will have some empty sets. So choose one block to be empty and count. etc...etc. If this is not useful, I suggest you try looking in this same forum under the tag "stirling-numbers" here: https://math.stackexchange.com/questions/tagged/stirling-numbers For sure you will find something. – Phicar Dec 24 '21 at 12:15