False statementConstruction: AI-adaptedVerification: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-27

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The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
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FALSE: $S (n, k)$ counts surjections $[n] \to [k]$

Statement

False claim: for all $n, k$ , the number $S (n, k)$ is the number of surjections $[n] \to [k]$ .

Facts & Assumptions

Given: The surjection formula and the second-kind inclusion-exclusion formula.

Proof

technique · direct

1.1given

For $n = 3$ and $k = 2$ , The Stirling numbers of the second kind are given by $S (n, k) = \frac{1}{k!} \sum_{i = 0}^{k} (- 1)^{i} (\binom{k}{i}) (k - i)^{n}$ gives $S (3, 2) = \frac{1}{2} (2^{3} - 2) = 3$ .

2.1step 1.1given

But 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 $ι$ gives the surjection count $2^{3} - 2 = 6$ . Since $3 \neq 6$ , $S (3, 2)$ does not count surjections themselves; it counts them only after dividing by $2!$ .

3.1step 2.1∎

Therefore the claim is false.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

21 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)

FALSE: S(n,k)S(n,k) counts surjections [n]→[k][n]→[k]