TheoremStatement: AI-adaptedProof: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-29

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The multinomial coefficient equals $n! / \prod_{i < m} k_{i}!$ , and $(x_{0} + \dots + x_{m - 1})^{n} = \sum ι ⁣ (\binom{n}{k}) \prod_{i < m} x_{i}^{k_{i}}$ in $R$

Statement

Let $m, n \in N$ and $k \in W (n, m)$ , that is $k : m \to N$ with $\sum_{i < m} k_{i} = n$ (The multinomial coefficient $(\binom{n}{k_{0}, \dots, k_{m - 1}})$ as the number of ordered partitions of an $n$ -set into blocks of prescribed sizes). Then:

The closed formula, in $N$ . $(\binom{n}{k}) \cdot \prod_{i < m} k_{i}! = n!,$ so in $R$ the quotient $ι (n!) / \prod_{i < m} ι (k_{i}!)$ is $ι (\binom{n}{k})$ , the canonical natural of a count.
The expansion, in $R$ . For $x : m \to R$ , $(\sum_{i < m} x_{i})^{n} = \sum_{k \in W (n, m)} ι (\binom{n}{k}) \prod_{i < m} x_{i}^{k_{i}},$ the outer sum being the sum over the finite index set $W (n, m)$ (The sum $\sum_{i \in S} a_{i}$ over a finite index set, and its product form) and the inner product the real finite product of Finite sums and finite products, by recursion.

As with the binomial theorem, the identity is stated in $R$ ; the commutative-ring version is a separate statement, to be made where rings exist. See the Remarks of The binomial theorem in $R$ : $(x + y)^{n} = \sum_{k < n + 1} ι ⁣ (\binom{n}{k}) x^{k} y^{n - k}$ .

Facts & Assumptions

Given: Naturals $m$ , $n$ , a tuple $k \in W (n, m)$ , a list $x : m \to R$ , and a finite set $A$ with $∣ A ∣ = n$ . For $m = σ (M)$ write $k^{'} : M \to N$ for the shifted tuple $k_{j}^{'} : = k_{σ (j)}$ .

[L1]

Induction (The principle of mathematical induction).

[L2]

Multinomial coefficients (The multinomial coefficient $(\binom{n}{k_{0}, \dots, k_{m - 1}})$ as the number of ordered partitions of an $n$ -set into blocks of prescribed sizes): $∣ B (A, k) ∣ = (\binom{∣ A ∣}{k})$ ; $B (A, k)$ and $W (n, m)$ are finite; $(\binom{0}{}) = 1$ for the empty tuple; and $W (n, 0)$ is ${()}$ for $n = 0$ and $\emptyset$ for $n \geq 1$ .

[L3]

The sum rule, in particular the splitting of a sum over a finite index set along a partition of that index set (The sum rule: a finite disjoint union is finite with $∣ A \cup B ∣ = ∣ A ∣ + ∣ B ∣$ and $∣ ⋃_{i \in I} A_{i} ∣ = \sum_{i \in I} ∣ A_{i} ∣$ , and a sum over a finite index set splits along a partition, clause 3), together with the reindexing and constant clauses of The sum $\sum_{i \in S} a_{i}$ over a finite index set, and its product form.

[L4]

The binomial closed formula $(\binom{n}{j}) j! (n - j)! = n!$ for $j \leq n$ ( $(\binom{n}{k}) k! (n - k)! = n!$ for $k \leq n$ ; hence $(\binom{n}{k}) k! = n^{\underline{k}}$ , the quotient $n! / (k! (n - k)!)$ is a natural number, and $(\binom{n}{k}) = (\binom{n}{n - k})$ ) and the binomial theorem (The binomial theorem in $R$ : $(x + y)^{n} = \sum_{k < n + 1} ι ⁣ (\binom{n}{k}) x^{k} y^{n - k}$ ).

[L5]

Finite sums and products: the recursion clauses in $N$ and in $R$ , the splitting clause at index $1$ , and the scaling and additivity clauses in $R$ (Finite sums and finite products of natural numbers, $\sum_{k < n} a_{k}$ and $\prod_{k < n} a_{k}$ in $N$ , Finite sums and finite products, by recursion, Laws of finite sums and finite products, Laws of finite sums and products in $N$ , and $ι (\sum_{k < n} a_{k}) = \sum_{k < n} ι (a_{k})$ ).

[L6]

Factorials: $j! \neq 0$ , and a product of nonzero naturals is nonzero; cancellation by a nonzero natural (The factorial $n!$ and the falling factorial $n^{\underline{k}}$ , defined by recursion in $N$ , Cancellation for multiplication by a nonzero factor, Multiplication is associative, Multiplication is commutative).

[L7]

$ι$ is additive, multiplicative, injective, and commutes with finite sums and products (clauses 0, 6, 7 of Laws of finite sums and products in $N$ , and $ι (\sum_{k < n} a_{k}) = \sum_{k < n} ι (a_{k})$ , The canonical natural $ι (n) = n \cdot 1_{F}$ of a field); $R$ is a field (Field); powers obey $a^{0} = 1$ , $a^{σ (j)} = a^{j} a$ (Integer powers $a^{m}$ ).

[L8]

Binomial coefficients and cardinality: $∣ [A]^{j} ∣ = (\binom{∣ A ∣}{j})$ (The set $[A]^{k}$ of $k$ -element subsets and the binomial coefficient $(\binom{n}{k}) : = ∣ [n]^{k} ∣$ ); transport (The cardinality $∣ A ∣$ of a finite set); a subset of a finite set is finite (A subset of a finite set is finite, with $∣ B ∣ \leq ∣ A ∣$ , and equality holds if and only if $B = A$ , The set $A^{B}$ of functions $B \to A$ between finite sets is finite, with $∣ A^{B} ∣ = ∣ A ∣^{∣ B ∣}$ ); the product rule (The product rule: $∣ A \times B ∣ = ∣ A ∣ ∣ B ∣$ , and $∣ \prod_{i < m} A_{i} ∣ = \prod_{i < m} ∣ A_{i} ∣$ ); two-sided inverses give bijections (Injection, surjection, bijection); every nonzero natural is a successor (Every nonzero natural number is a successor); $j + (n - j) = n$ for $j \leq n$ (Order on the natural numbers, Addition is cancellative).

Proof

technique · induction

1.1

Notation for both inductions. For $m = σ (M)$ and $k \in W (n, σ (M))$ , splitting the sum at index $1$ ([L5]) gives $n = \sum_{i < σ (M)} k_{i} = k_{0} + \sum_{j < M} k_{j}^{'}$ , so $k^{'} \in W (n - k_{0}, M)$ and $k_{0} \leq n$ ; the same splitting for products gives $\prod_{i < σ (M)} k_{i}! = k_{0}! \cdot \prod_{j < M} k_{j}^{'}!$ .

givenL5L8

1.2

Base case of clause 1, at $m = 0$ . Then $W (n, 0)$ is nonempty only for $n = 0$ , and there $(\binom{0}{}) = 1$ , the empty product of factorials is $1$ and $0! = 1$ , so the identity reads $1 \cdot 1 = 1$ .

baseL2L5L6

1.3

Inductive hypothesis for clause 1: fix $M$ and assume $(\binom{N}{k^{'}}) \prod_{j < M} k_{j}^{'}! = N!$ for every $N$ and every $k^{'} \in W (N, M)$ .

1.4

Base case of clause 2, at $m = 0$ . The left-hand side is $(\sum_{i < 0} x_{i})^{n} = 0^{n}$ . If $n = 0$ this is $1$ , and the right-hand side is the single term $ι (\binom{0}{}) \prod_{i < 0} x_{i}^{k_{i}} = 1 \cdot 1 = 1$ . If $n \geq 1$ then $0^{n} = 0$ and $W (n, 0) = \emptyset$ , so the right-hand side is an empty sum, equal to $0$ .

L2L5L7

2.1

Inductive step for clause 1. Let $m = σ (M)$ , $k \in W (n, σ (M))$ , and let $A$ be finite with $∣ A ∣ = n$ . The map $c \mapsto (c^{- 1} [{0}], c^{''})$ , where $c^{''} (a)$ is the unique $i$ with $c (a) = σ (i)$ for $a \notin c^{- 1} [{0}]$ , sends $B (A, k)$ to the set of pairs $(S, c^{''})$ with $S \in [A]^{k_{0}}$ and $c^{''} \in B (A ∖ S, k^{'})$ ; it is well defined because $c (a) \neq 0$ off the first fibre, every nonzero natural is a unique successor, and $∣ c^{'' - 1} [{j}] ∣ = ∣ c^{- 1} [{σ (j)}] ∣ = k_{j}^{'}$ . Its two-sided inverse sends $(S, c^{''})$ to the colouring equal to $0$ on $S$ and to $σ (c^{''} (a))$ off $S$ . The pairs form the union of the pairwise disjoint sets ${S} \times B (A ∖ S, k^{'})$ indexed by $S \in [A]^{k_{0}}$ , and $∣ A ∖ S ∣ = n - k_{0}$ by [L3], so each has $(\binom{n - k_{0}}{k^{'}})$ elements. Hence $(\binom{n}{k}) = (\binom{n}{k_{0}}) (\binom{n - k_{0}}{k^{'}})$ by [L3] and [L8]. Multiplying by $\prod_{i < σ (M)} k_{i}! = k_{0}! \prod_{j < M} k_{j}^{'}!$ and using the hypothesis of step 1.3 at $N = n - k_{0}$ gives $(\binom{n}{k}) \prod_{i < σ (M)} k_{i}! = (\binom{n}{k_{0}}) k_{0}! \cdot ((\binom{n - k_{0}}{k^{'}}) \prod_{j < M} k_{j}^{'}!) = (\binom{n}{k_{0}}) k_{0}! (n - k_{0})! = n!$ by [L4], since $k_{0} \leq n$ .

step 1.1step 1.3L2L3L4L6L8construct

3.1

Clause 1 holds for every $m$ , by step 1.2, step 2.1 and induction. The real form follows: applying $ι$ gives $ι (\binom{n}{k}) \prod_{i < m} ι (k_{i}!) = ι (n!)$ , and each $ι (k_{i}!)$ is nonzero by [L6] and [L7], so the product is invertible in $R$ .

step 1.2step 2.1L1L6L7

4.1

Inductive step for clause 2. Assume clause 2 at $M$ , for every $n$ and every list of length $M$ . Let $x : σ (M) \to R$ , put $y : = \sum_{i < M} x_{i}$ and $z : = x_{M}$ , so $\sum_{i < σ (M)} x_{i} = y + z$ by [L5]. The map $Φ (j, κ) : = \hat{κ}$ , where $\hat{κ}$ restricts to $κ$ on $M$ and ${\hat{κ}}_{M} : = n - j$ , is a bijection from the disjoint union of the sets ${j} \times W (j, M)$ , $j < σ (n)$ , onto $W (n, σ (M))$ : it lands there because $\sum_{i < σ (M)} {\hat{κ}}_{i} = j + (n - j) = n$ , and its inverse sends $\hat{κ}$ to the pair $(\sum_{i < M} {\hat{κ}}_{i}, \hat{κ} ↾ M)$ , the two constructions being mutually inverse by [L8]. Moreover $ι (\binom{n}{j}) ι (\binom{j}{κ}) = ι (\binom{n}{\hat{κ}})$ : by step 3.1 and [L4], both $(\binom{n}{\hat{κ}})$ and $(\binom{n}{j}) (\binom{j}{κ})$ become $n!$ after multiplication by the nonzero natural $(\prod_{i < M} κ_{i}!) (n - j)!$ , so they are equal by cancellation. And $\prod_{i < M} x_{i}^{κ_{i}} \cdot z^{n - j} = \prod_{i < σ (M)} x_{i}^{{\hat{κ}}_{i}}$ by the product recursion clause. Now [L4] gives $(y + z)^{n} = \sum_{j < σ (n)} ι (\binom{n}{j}) y^{j} z^{n - j}$ ; substituting the inductive hypothesis for $y^{j}$ , distributing the scalar $ι (\binom{n}{j}) z^{n - j}$ over the inner sum by the scaling clause of [L5], and then applying [L3] to the partition of $W (n, σ (M))$ into the images of the sets $W (j, M)$ under $Φ$ yields $(\sum_{i < σ (M)} x_{i})^{n} = \sum_{\hat{κ} \in W (n, σ (M))} ι (\binom{n}{\hat{κ}}) \prod_{i < σ (M)} x_{i}^{{\hat{κ}}_{i}}$ .

step 1.4step 3.1assume-hypL2L3L4L5L6L7L8construct

5.1

By step 1.4, step 4.1 and induction, clause 2 holds for every $m \in N$ , every $n$ and every list $x$ .

step 1.4step 4.1L1

6.1

Clause 1 is step 3.1 and clause 2 is step 5.1.

step 3.1step 5.1discharge-induction∎

Remarks

The index set of the outer sum has to be finite, and it is. It is $W (n, m)$ , which The multinomial coefficient $(\binom{n}{k_{0}, \dots, k_{m - 1}})$ as the number of ordered partitions of an $n$ -set into blocks of prescribed sizes shows finite by injecting it into the set of functions $m \to σ (n)$ . Without that the outer sum would not be defined, which is the reason The sum $\sum_{i \in S} a_{i}$ over a finite index set, and its product form exists.
The small cases are computed, not waved at. At $m = 0$ the left-hand side is $0^{n}$ and the right-hand side is an empty sum or a single term, and the two match only because $0^{0} = 1$ . At $m = 1$ the only tuple is $k = (n)$ , the coefficient is $(\binom{n}{n}) = 1$ by clause 1, and the identity reads $x_{0}^{n} = x_{0}^{n}$ .
Clause 1 is again an identity between natural numbers, so integrality of $n! / \prod k_{i}!$ is free; the quotient form is a consequence obtained through $ι$ , exactly as for the binomial coefficient.

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Sources

Multinomial theorem (Wikipedia) (standard reference, not scraped)
Binomial theorem (Wikipedia) (standard reference, not scraped)
R. Stanley, Enumerative Combinatorics, Vol. 1, Ch. 1 (standard reference, not scraped)

The multinomial coefficient equals n!/∏i<mki!n!/∏i<m​ki​!, and (x0+⋯+xm−1)n=∑ι ⁣(nk)∏i<mxiki(x0​+⋯+xm−1​)n=∑ι(kn​)∏i<m​xiki​​ in RR