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All nine derangements of a four-element set listed, and the count checked against the formula and both recurrences

Example

Take $A : = 4 = {0, 1, 2, 3}$ and write a bijection $f : A \to A$ as the tuple $(f (0), f (1), f (2), f (3))$ . The derangements of $A$ (The derangement number $D_{n}$ : the number of bijections of an $n$ -element set with no fixed point) are exactly

$\begin{array}{lll} (1, 0, 3, 2) & (1, 2, 3, 0) & (1, 3, 0, 2) \\ (2, 0, 3, 1) & (2, 3, 0, 1) & (2, 3, 1, 0) \\ (3, 0, 1, 2) & (3, 2, 0, 1) & (3, 2, 1, 0) \end{array}$

so $D_{4} = 9$ . Each tuple lists four distinct values, hence is a bijection, and no entry equals its position.

Against the formula. $ι (D_{n}) = ι (n!) \sum_{i < n + 1} (- 1)^{i} / ι (i!)$ , with the term at $i = 0$ equal to $1$ and $D_{0} = 1$ gives

$ι (D_{4}) = ι (4!) \sum_{i < 5} \frac{(- 1)^{i}}{ι (i!)} = 24 (1 - 1 + \frac{1}{2} - \frac{1}{6} + \frac{1}{24}) = 24 \cdot \frac{9}{24} = 9.$

Against the two recurrences. The earlier values are $D_{0} = 1$ , $D_{1} = 0$ , $D_{2} = 1$ and, by the first recurrence, $D_{3} = 3 D_{2} + (- 1)^{3} = 3 - 1 = 2$ . Then $ι (D_{n}) = ι (n) ι (D_{n - 1}) + (- 1)^{n}$ for $n \geq 1$ , and $D_{n} = (n - 1) (D_{n - 1} + D_{n - 2})$ for $n \geq 2$ gives $ι (D_{4}) = ι (4) ι (D_{3}) + (- 1)^{4} = 4 \cdot 2 + 1 = 9$ from its first clause, and $D_{4} = 3 (D_{3} + D_{2}) = 3 (2 + 1) = 9$ from its second.

Facts & Assumptions

Given: $A = 4 = {0, 1, 2, 3}$ , the tuple notation above, and the canonical natural $ι$ (The canonical natural $ι (n) = n \cdot 1_{F}$ of a field).

[L1]

$∣ A ∣ = ∣ 4 ∣ = 4$ (The cardinality $∣ A ∣$ of a finite set, clause (a)), so $∣ Bij (A) ∣ = 4! = 24$ (A finite set $A$ with $∣ A ∣ = n$ has exactly $n!$ bijections onto itself, and $n!$ bijections onto any set of the same cardinality, The factorial $n!$ and the falling factorial $n^{\underline{k}}$ , defined by recursion in $N$ ); and a tuple of four distinct entries drawn from $A$ is a bijection $A \to A$ , a function on a finite set being injective exactly when it is a bijection (Injection, surjection, bijection, A subset of a finite set is finite, with $∣ B ∣ \leq ∣ A ∣$ , and equality holds if and only if $B = A$ , clause 4).

[L2]

For every finite set $B$ , $Der (B)$ is the set of bijections $f : B \to B$ with $f (b) \neq b$ for every $b \in B$ , and $∣ Der (B) ∣ = D_{∣ B ∣}$ ; hence $D_{4} = ∣ Der (A) ∣$ because $∣ A ∣ = 4$ (The derangement number $D_{n}$ : the number of bijections of an $n$ -element set with no fixed point, The cardinality $∣ A ∣$ of a finite set, clause (a)). The values $D_{0} = 1$ , $D_{1} = 0$ and $D_{2} = 1$ are also recorded in the cited definition.

[L3]

For every $n \in N$ , $ι (D_{n}) = ι (n!) \sum_{i < n + 1} (- 1)^{i} / ι (i!)$ . For $n \geq 1$ , $ι (D_{n}) = ι (n) ι (D_{n - 1}) + (- 1)^{n}$ ; and for $n \geq 2$ , $D_{n} = (n - 1) (D_{n - 1} + D_{n - 2})$ ( $ι (D_{n}) = ι (n!) \sum_{i < n + 1} (- 1)^{i} / ι (i!)$ , with the term at $i = 0$ equal to $1$ and $D_{0} = 1$ , $ι (D_{n}) = ι (n) ι (D_{n - 1}) + (- 1)^{n}$ for $n \geq 1$ , and $D_{n} = (n - 1) (D_{n - 1} + D_{n - 2})$ for $n \geq 2$ ).

[L4]

Factorials: $0! = 1$ , $1! = 1$ , $2! = 2$ , $3! = 6$ , $4! = 24$ (The factorial $n!$ and the falling factorial $n^{\underline{k}}$ , defined by recursion in $N$ ).

[L5]

Real finite sums and the arithmetic of $R$ : recursion clause, additivity and scaling (Finite sums and finite products, by recursion, Laws of finite sums and finite products, Ordered field); and $(- 1)^{0} = 1$ , $(- 1)^{p + 1} = - (- 1)^{p}$ (Integer powers $a^{m}$ ).

[L6]

Cardinality of a listed set with distinct entries (The cardinality $∣ A ∣$ of a finite set, clauses (a) and (c)).

[L7]

The canonical natural $ι : N \to R$ is additive, multiplicative and injective (Laws of finite sums and products in $N$ , and $ι (\sum_{k < n} a_{k}) = \sum_{k < n} ι (a_{k})$ , clauses 0 and 7).

Verification

technique · cases

1.1

Each of the nine listed tuples has four distinct entries and no entry equal to its position, so each is a derangement of $A$ by [L1] and [L2]; and the nine tuples are pairwise distinct, so the listed set has nine elements by [L6].

givenL1L2L6

1.2

The enumeration is complete, and the cases are indexed by the value $f (0)$ , which is $1$ , $2$ or $3$ since $f (0) \neq 0$ .

givenL2

2.1

Case $f (0) = 1$ . Then $f (1) \in {0, 2, 3}$ . If $f (1) = 0$ the remaining values ${2, 3}$ go to positions $2$ and $3$ with $f (2) \neq 2$ , forcing $(1, 0, 3, 2)$ . If $f (1) = 2$ the remaining values are ${0, 3}$ with $f (3) \neq 3$ , forcing $(1, 2, 3, 0)$ . If $f (1) = 3$ the remaining values are ${0, 2}$ with $f (2) \neq 2$ , forcing $(1, 3, 0, 2)$ .

assume-case onestep 1.2L2

2.2

Case $f (0) = 2$ . Then $f (1) \in {0, 3}$ . If $f (1) = 0$ the remaining values ${1, 3}$ go to positions $2$ and $3$ with $f (3) \neq 3$ , forcing $(2, 0, 3, 1)$ . If $f (1) = 3$ the remaining values are ${0, 1}$ and neither placement is excluded, giving $(2, 3, 0, 1)$ and $(2, 3, 1, 0)$ .

assume-case twostep 1.2L2

2.3

Case $f (0) = 3$ . Then $f (1) \in {0, 2}$ . If $f (1) = 0$ the remaining values ${1, 2}$ go to positions $2$ and $3$ with $f (2) \neq 2$ , forcing $(3, 0, 1, 2)$ . If $f (1) = 2$ the remaining values are ${0, 1}$ and neither placement is excluded, giving $(3, 2, 0, 1)$ and $(3, 2, 1, 0)$ .

assume-case threestep 1.2L2

3.1

The three cases are exhaustive and produce exactly the nine listed tuples, so $D_{4} = 9$ .

step 1.1step 2.1step 2.2step 2.3cases-exhaustiveL2

4.1

Against the formula. By [L3] and [L4], $ι (D_{4}) = ι (24) (1 - 1 + 1 / 2 - 1 / 6 + 1 / 24)$ ; the bracket is $9 / 24$ by [L5], so $ι (D_{4}) = 9$ , matching step 3.1.

step 3.1L3L4L5

4.2

Against the recurrences. By [L2] and the first clause of [L3], $ι (D_{3}) = ι (3) ι (D_{2}) + (- 1)^{3} = ι (3) - ι (1) = ι (2)$ , so $D_{3} = 2$ by injectivity of $ι$ . Then $ι (D_{4}) = ι (4) ι (D_{3}) + (- 1)^{4} = ι (8) + ι (1) = ι (9)$ ; by the second clause, $D_{4} = 3 (D_{3} + D_{2}) = 3 \cdot 3 = 9$ . Both match step 3.1.

step 3.1L2L3L5L7

5.1

The list, the formula and the two recurrences therefore all give $D_{4} = 9$ .

step 3.1step 4.1step 4.2∎

Remarks

The case analysis is on $f (0)$ and then on $f (1)$ , and the remaining two positions are then forced or free according to whether the two leftover values can be placed without creating a fixed point. Where exactly one placement avoids a fixed point the tuple is determined; where both do, the subcase splits. The branching pattern therefore differs between the cases even though each contributes the same number of derangements.
The recurrences are checked at their first legal indices too. The first recurrence is used at $n = 3$ and at $n = 4$ , both at least $1$ ; the second at $n = 4$ , which is at least $2$ . Neither is evaluated where its hypothesis fails.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

64 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

Derangement (Wikipedia) (standard reference, not scraped)
Rencontres numbers (Wikipedia) (standard reference, not scraped)
Principle of Inclusion and Exclusion (Open Math Books) (standard reference, not scraped)