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How many different ways do there exist to split the set $A=\{1,2,...,n\}$ into at most $m$ non-empty disjoint subsets, whose union is $A$.

For example, if $m=3$ then we have the following:

$n=1: \quad$ there is only 1 such subset;

$n=2: \quad$ we have the following possible splittings:

$\quad\quad\quad\quad\big\{\{1\}, \{2\}\big\}, \big\{\{1,2\}\big\}$ - $2$ in total;

$n=3: \quad$ we have the following possible splittings:

$\quad\quad\quad\quad\big\{\{1\}, \{2\},\{3\}\big\}, \big\{\{1,2\}, \{3\}\big\}, \big\{\{1,3\}, \{2\}\big\}, \big\{\{1\}, \{2,3\}\big\}, \big\{\{1,2,3\}\big\}$ - $5$ in total.

samb
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1 Answers1

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Let $S(n,k)$ be the so-called Stirling number of the second kind which is the number of ways to partition a set of $n$ objects into exactly $k$ non-empty subsets. Hence the number of ways to split the set $\{1,2,...,n\}$ into at most $m$ non-empty disjoint subsets, whose union is $A$ is $$\sum_{k=0}^m S(n,k)=\sum_{k=0}^m\frac{1}{k!}\sum_{j=0}^k (-1)^j\binom{k}{j}(k-j)^n.$$

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Robert Z
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  • For completeness sake, you might consider indexing starting from $k=0$ just in case the OP allows for the case where $n=0$, noting that $S(0,0)=1$ referring to how the empty partition is a valid partition of the empty set. It wouldn't affect the result in other situations since $S(n,0)=0$ when $n>0$. Otherwise, if we were to restrict this to $n\geq 1$, this is of course fine. – JMoravitz Aug 03 '20 at 13:22
  • @JMoravitz Thanks for pointing out. – Robert Z Aug 03 '20 at 18:37