TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-27

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The Stirling numbers of the second kind are given by $S (n, k) = \frac{1}{k!} \sum_{i = 0}^{k} (- 1)^{i} (\binom{k}{i}) (k - i)^{n}$

Statement

For all $n, k \in N$ ,

$S (n, k) = \frac{1}{k!} \sum_{i = 0}^{k} (- 1)^{i} (\binom{k}{i}) (k - i)^{n} .$

Facts & Assumptions

Given: The surjection-counting formula of The number of surjections from an $n$ -element set onto a $k$ -element set is $\sum_{i < k + 1} (- 1)^{i} (\binom{k}{i}) (k - i)^{n}$ , read in $R$ through $ι$ and the second-kind definition of The Stirling numbers of the second kind and the Bell numbers.

Proof

technique · direct

1.1given

A partition of $[n]$ into exactly $k$ blocks becomes a surjection $[n] \to [k]$ once the $k$ blocks are labelled by the $k$ elements of $[k]$ . Conversely, the fibres of a surjection $[n] \to [k]$ form a partition of $[n]$ into exactly $k$ nonempty blocks. Thus the number of surjections $[n] \to [k]$ is $k! S (n, k)$ .

1.2given

By The number of surjections from an $n$ -element set onto a $k$ -element set is $\sum_{i < k + 1} (- 1)^{i} (\binom{k}{i}) (k - i)^{n}$ , read in $R$ through $ι$ , that same number equals $\sum_{i = 0}^{k} (- 1)^{i} (\binom{k}{i}) (k - i)^{n}$ . Therefore $k! S (n, k) = \sum_{i = 0}^{k} (- 1)^{i} (\binom{k}{i}) (k - i)^{n} .$

2.1step 1.1step 1.2∎

Dividing by the nonzero factorial $k!$ from The factorial $n!$ and the falling factorial $n^{\underline{k}}$ , defined by recursion in $N$ gives the displayed formula.

Depends on

Used by

The second-kind inclusion-exclusion formula does not count permutations by cycles Counterexample
S(4,2) computed by recurrence, inclusion-exclusion, and direct listing Example
FALSE: S(n,k) counts surjections [n]→[k] False statement
The twelvefold way Theorem

Dependency tree · two levels

33 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Andrew Lin, 18.212 Algebraic Combinatorics, Lecture 11 (standard reference, not scraped)
Herbert S. Wilf, generatingfunctionology, 2nd ed., §3.6 (standard reference, not scraped)

The Stirling numbers of the second kind are given by S(n,k)=1k!∑i=0k(−1)i(ki)(k−i)nS(n,k)=k!1​∑i=0k​(−1)i(ik​)(k−i)n