The most elegant way to generate permutations in SQL server

Question

Given a the following table:

Index | Element
---------------
  1   |    A
  2   |    B
  3   |    C
  4   |    D

We want to generate all th

Answer 1

Current solution using a recursive CTE.

-- The base elements
Declare @Number Table( Element varchar(MAX), Id varchar(MAX) )
Insert Into @Number Values ( 'A', '01')
Insert Into @Number Values ( 'B', '02')
Insert Into @Number Values ( 'C', '03')
Insert Into @Number Values ( 'D', '04')

-- Number of elements
Declare @ElementsNumber int
Select  @ElementsNumber = COUNT(*)
From    @Number;



-- Permute!
With Permutations(   Permutation,   -- The permutation generated
                     Ids,            -- Which elements where used in the permutation
                     Depth )         -- The permutation length
As
(
    Select  Element,
            Id + ';',
            Depth = 1
    From    @Number
    Union All
    Select  Permutation + ' ' + Element,
            Ids + Id + ';',
            Depth = Depth + 1
    From    Permutations,
            @Number
    Where   Depth < @ElementsNumber And -- Generate only the required permutation number
            Ids Not like '%' + Id + ';%' -- Do not repeat elements in the permutation (this is the reason why we need the 'Ids' column) 
)
Select  Permutation
From    Permutations
Where   Depth = @ElementsNumber