Derangements on fintypes #
This file contains lemmas that describe the cardinality of derangements α
when α
is a fintype.
Main definitions #
card_derangements_invariant
: A lemma stating that the number of derangements on a typeα
depends only on the cardinality ofα
.numDerangements n
: The number of derangements on an n-element set, defined in a computation- friendly way.card_derangements_eq_numDerangements
: Proof thatnumDerangements
really does compute the number of derangements.numDerangements_sum
: A lemma giving an expression fornumDerangements n
in terms of factorials.
Equations
- instDecidablePredPermDerangements x = Fintype.decidableForallFintype
instance
instFintypeElemPermDerangements
{α : Type u_1}
[DecidableEq α]
[Fintype α]
:
Fintype ↑(derangements α)
Equations
- instFintypeElemPermDerangements = Subtype.fintype fun (x : Equiv.Perm α) => ∀ (x_1 : α), ¬x x_1 = x_1
theorem
card_derangements_invariant
{α : Type u_2}
{β : Type u_3}
[Fintype α]
[DecidableEq α]
[Fintype β]
[DecidableEq β]
(h : Fintype.card α = Fintype.card β)
:
Fintype.card ↑(derangements α) = Fintype.card ↑(derangements β)
theorem
card_derangements_fin_add_two
(n : ℕ)
:
Fintype.card ↑(derangements (Fin (n + 2))) = (n + 1) * Fintype.card ↑(derangements (Fin n)) + (n + 1) * Fintype.card ↑(derangements (Fin (n + 1)))
The number of derangements of an n
-element set.
Equations
- numDerangements 0 = 1
- numDerangements 1 = 0
- numDerangements n.succ.succ = (n + 1) * (numDerangements n + numDerangements (n + 1))
Instances For
theorem
numDerangements_add_two
(n : ℕ)
:
numDerangements (n + 2) = (n + 1) * (numDerangements n + numDerangements (n + 1))
theorem
numDerangements_succ
(n : ℕ)
:
↑(numDerangements (n + 1)) = (↑n + 1) * ↑(numDerangements n) - (-1) ^ n
theorem
card_derangements_fin_eq_numDerangements
{n : ℕ}
:
Fintype.card ↑(derangements (Fin n)) = numDerangements n
theorem
card_derangements_eq_numDerangements
(α : Type u_2)
[Fintype α]
[DecidableEq α]
:
Fintype.card ↑(derangements α) = numDerangements (Fintype.card α)
theorem
numDerangements_sum
(n : ℕ)
:
↑(numDerangements n) = ∑ k ∈ Finset.range (n + 1), (-1) ^ k * ↑((k + 1).ascFactorial (n - k))