ExampleConstruction: AI-adaptedVerification: AI-generatedprecheck passaudited 2026-08-01

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$τ (K_{4}) = 16$ by direct Prüfer enumeration and Cayley's formula

Statement

The complete graph $K_{4}$ has $16$ spanning trees.

Facts & Assumptions

Given: $K_{4}$ on the label set $4$ .

[L1]

Its spanning trees correspond bijectively to words of length $2$ on $4$ (Prüfer encoding and decoding are inverse bijections between labelled trees on $n \geq 2$ vertices and words of length $n - 2$ on their labels).

[L2]

Cayley's formula gives $τ (K_{n}) = n^{n - 2}$ (Cayley's formula: $τ (K_{n}) = n^{n - 2}$ for $n \geq 2$ , with $τ (K_{1}) = 1$ and $τ (K_{0}) = 0$ ).

[F1]

$τ$ counts spanning trees (The spanning-tree number $τ (G)$ ).

Verification

technique · direct

1.1

There are $4$ choices for each of the two positions of a Prüfer word, hence $4^{2} = 16$ words and therefore $16$ spanning trees.

L1F1

2.1

Independently, Cayley's formula gives $τ (K_{4}) = 4^{4 - 2} = 16$ , agreeing with the enumeration.

L2∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

13 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

ISI Bangalore discrete mathematics notes, Trees and Cayley’s theorem (standard reference, not scraped)

τ(K4)=16τ(K4​)=16 by direct Prüfer enumeration and Cayley's formula