ExampleConstruction: AI-adaptedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passverified 2026-08-02 (claude-opus-5)

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Arrangements of a word with repeated letters, counted by the multinomial coefficient

Example

Take the eleven-letter word $\mathrm{MISSISSIPPI}$ , over the four-letter alphabet $\{\mathrm{M},\mathrm{I},\mathrm{S},\mathrm{P}\}$ , in which $\mathrm{M}$ occurs once, $\mathrm{I}$ four times, $\mathrm{S}$ four times and $\mathrm{P}$ twice. The number of distinct arrangements of its letters is

$\binom{11}{1,4,4,2} = \frac{11!}{1!\,4!\,4!\,2!} = \frac{39916800}{1152} = 34650 .$

The modelling step is the mathematics. An arrangement is a function from the set of eleven positions to the four-letter alphabet whose fibre over each letter has the prescribed size; that is literally an element of $\mathcal{B}(A,k)$ in 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, with $A$ the set of positions, $m = 4$ and $k = (1,4,4,2)$ .

Facts & Assumptions

Given: The position set $A$ with $\lvert A\rvert = 11$ , the alphabet identified with $4 = \{0,1,2,3\}$ by $0 \mapsto \mathrm{M}$ , $1 \mapsto \mathrm{I}$ , $2 \mapsto \mathrm{S}$ , $3 \mapsto \mathrm{P}$ , and the tuple $k = (1,4,4,2)$ ; the factorials $1! = 1$ , $2! = 2$ , $4! = 24$ and $11! = 39916800$ (The factorial $n!$ and the falling factorial $n^{\underline{k}}$ , defined by recursion in $\mathbb{N}$ ).

[L1]

$\mathcal{B}(A,k)$ is the set of $c : A \to m$ with $\lvert c^{-1}[\{i\}]\rvert = k_i$ for every $i<m$ ; it is nonempty only if $\sum_{i<m}k_i = \lvert A\rvert$ , and its cardinality is $\binom{\lvert A\rvert}{k}$ (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, The cardinality $\lvert A\rvert$ of a finite set).

[L2]

$\binom{n}{k}\prod_{i<m}k_i! = n!$ (The multinomial coefficient equals $n!/\prod_{i<m} k_i!$ , and $(x_0+\dots+x_{m-1})^{n} = \sum \iota\!\binom{n}{k}\prod_{i<m} x_i^{k_i}$ in $\mathbb{R}$ , clause 1).

[L3]

Finite sums and products in $\mathbb{N}$ and cancellation by a nonzero natural (Finite sums and finite products of natural numbers, $\sum_{k<n} a_k$ and $\prod_{k<n} a_k$ in $\mathbb{N}$ , Cancellation for multiplication by a nonzero factor).

Verification

technique · direct

1.1

The modelling. An arrangement of the letters of $\mathrm{MISSISSIPPI}$ is a function $c$ assigning to each of the eleven positions one of the four letters, subject to the letter multiplicities; that is, $\lvert c^{-1}[\{0\}]\rvert = 1$ , $\lvert c^{-1}[\{1\}]\rvert = 4$ , $\lvert c^{-1}[\{2\}]\rvert = 4$ and $\lvert c^{-1}[\{3\}]\rvert = 2$ . So the set of arrangements is exactly $\mathcal{B}(A,k)$ with $k = (1,4,4,2)$ .

givenL1

2.1

The hypothesis is satisfied, and it must be checked before the coefficient is written down: $\sum_{i<4}k_i = 1+4+4+2 = 11 = \lvert A\rvert$ , so $k \in \mathcal{W}(11,4)$ and $\binom{11}{1,4,4,2}$ is defined.

step 1.1L1L3

3.1

The value. By [L2], $\binom{11}{1,4,4,2}\cdot(1!\cdot 4!\cdot 4!\cdot 2!) = 11!$ , that is $\binom{11}{1,4,4,2}\cdot(1\cdot 24\cdot 24\cdot 2) = 39916800$ ; the product of factorials is $1152$ , and $1152 \cdot 34650 = 39916800$ , so cancellation by the nonzero factor $1152$ gives $\binom{11}{1,4,4,2} = 34650$ .

step 2.1L2L3

4.1

Hence the word has exactly $34650$ distinct arrangements, this being $\lvert\mathcal{B}(A,k)\rvert$ by [L1].

step 1.1step 3.1L1∎

Remarks

Why the letters are identified with $\{0,1,2,3\}$ . 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 takes the colour set to be a natural number, so the alphabet has to be presented as one; any bijection will do, and the count does not depend on which. A different identification permutes the tuple $k$ , and by clause 1 of The multinomial coefficient equals $n!/\prod_{i<m} k_i!$ , and $(x_0+\dots+x_{m-1})^{n} = \sum \iota\!\binom{n}{k}\prod_{i<m} x_i^{k_i}$ in $\mathbb{R}$ the coefficient is determined by $n$ together with the product $\prod_{i<m} k_i!$ , which a permutation of the parts leaves unchanged. The invariance clause of 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 is a different statement: it says the count depends only on $\lvert A\rvert$ , that is, only on the domain up to bijection.
The check in step 2.1 is not a formality. If the multiplicities did not sum to the length, $\mathcal{B}(A,k)$ would be empty and the symbol $\binom{11}{k}$ would not be defined at all.

Depends on

Used by

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Sources

Anagram (Wikipedia) (standard reference, not scraped)
Multinomial theorem (Wikipedia) (standard reference, not scraped)
Permutation (Wikipedia) (standard reference, not scraped)