ExampleConstruction: AI-generatedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (z-ai/glm-5.2)audited 2026-07-29

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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.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Choosing a committee: $\binom{10}{3} = 120$ , and the ordered count $10^{\underline{3}} = 720$

Example

Let $P$ be a set with $\lvert P\rvert = 10$ . Two different sets are counted, and naming which is which is the whole discipline of the example.

The unordered selections of three members of $P$ are the elements of $[P]^{3}$ , and there are $\binom{10}{3} = 120$ of them.
The ordered selections of three distinct members, that is the injections $3 \to P$ , number $10^{\underline{3}} = 720$ .

The ratio of the two counts is $3! = 6$ , which is clause 1 of $\binom{n}{k}\,k!\,(n-k)! = n!$ for $k \le n$ ; hence $\binom{n}{k}\,k! = n^{\underline{k}}$ , the quotient $n!/(k!(n-k)!)$ is a natural number, and $\binom{n}{k} = \binom{n}{n-k}$ seen concretely: each $3$ -element subset arises from exactly $3!$ ordered selections.

Facts & Assumptions

Given: A set $P$ with $\lvert P\rvert = 10$ , and the factorials $3! = 6$ , $7! = 5040$ , $10! = 3628800$ (The factorial $n!$ and the falling factorial $n^{\underline{k}}$ , defined by recursion in $\mathbb{N}$ ).

[L1]

$\lvert [P]^{k}\rvert = \binom{\lvert P\rvert}{k}$ (The set $[A]^{k}$ of $k$ -element subsets and the binomial coefficient $\binom{n}{k} := \lvert [n]^{k}\rvert$ , The cardinality $\lvert A\rvert$ of a finite set).

[L2]

$\binom{n}{k}\,k!\,(n-k)! = n!$ for $k \le n$ , and $\binom{n}{k}\,k! = n^{\underline{k}}$ ( $\binom{n}{k}\,k!\,(n-k)! = n!$ for $k \le n$ ; hence $\binom{n}{k}\,k! = n^{\underline{k}}$ , the quotient $n!/(k!(n-k)!)$ is a natural number, and $\binom{n}{k} = \binom{n}{n-k}$ , clause 1).

[L3]

The number of injections of a $k$ -element set into an $n$ -element set is $n^{\underline{k}}$ (The number of injections from a $k$ -element set into an $n$ -element set is $n^{\underline{k}}$ ).

[L4]

$n^{\underline{0}} = 1$ and $n^{\underline{\sigma(k)}} = n^{\underline{k}}(n-k)$ (The factorial $n!$ and the falling factorial $n^{\underline{k}}$ , defined by recursion in $\mathbb{N}$ ).

Verification

technique · direct

1.1

The two sets are $[P]^{3}$ , whose elements are the $3$ -element subsets of $P$ , and $\operatorname{Inj}(3,P)$ , whose elements are the injective functions from $3 = \{0,1,2\}$ into $P$ . They are different sets, and each count below is stated for the set it counts.

givenL1L3

2.1

The unordered count. By [L2] with $n = 10$ , $k = 3$ , $\binom{10}{3}\cdot 3!\cdot 7! = 10!$ , that is $\binom{10}{3}\cdot 6 \cdot 5040 = 3628800$ , so $\binom{10}{3}\cdot 30240 = 3628800$ and $\binom{10}{3} = 120$ . By [L1] the set $[P]^{3}$ has $120$ elements.

step 1.1L1L2

2.2

The ordered count. By [L4], $10^{\underline{1}} = 1 \cdot 10 = 10$ , $10^{\underline{2}} = 10\cdot 9 = 90$ and $10^{\underline{3}} = 90 \cdot 8 = 720$ . By [L3] the set $\operatorname{Inj}(3,P)$ has $720$ elements.

step 1.1L3L4

3.1

The two counts are related as clause 1 of [L2] says: $\binom{10}{3}\cdot 3! = 120 \cdot 6 = 720 = 10^{\underline{3}}$ . Each $3$ -element subset of $P$ is the image of exactly $3! = 6$ injections $3 \to P$ , so passing from the ordered to the unordered count divides by $6$ .

step 2.1step 2.2L2∎

Remarks

The standard error is to count one set and name the other. "How many ways can a committee of three be chosen from ten people" is the count of $[P]^{3}$ only if the committee is unordered; if the three roles are distinguished it is the count of $\operatorname{Inj}(3,P)$ . The two differ by a factor of $6$ , and no computation can decide which was meant.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 64 results over 21 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

Combination (Wikipedia) (standard reference, not scraped)
Binomial coefficient (Wikipedia) (standard reference, not scraped)
Falling and rising factorials (Wikipedia) (standard reference, not scraped)

Choosing a committee: (103)=120\binom{10}{3} = 120(310​)=120, and the ordered count 103‾=72010^{\underline{3}} = 720103​=720