Permutations

[Jez1]

Here is some code I got a while ago, to tell me all the different combinations of 1-9, then save to a txt file.

Option Explicit

'Must Declare all variables, I think that it is good practice
Option Base 1 'For simplicity, make array index start at 1 To n


Sub Perm(A() As String * 1, m, n As Integer)
    Dim i As Integer
    Dim Temp As String * 1

    If m = 1 Then
        PrintPerm A, n 'Print out the permutations
    Else
        For i = 1 To m
            Temp = A(i)
            A(i) = A(m) 'Exchange A(i) and A(m), could have made a Sub To Do this
            A(m) = Temp
            Perm A, m - 1, n 'Recursive Function call
            Temp = A(m)
            A(m) = A(i) 'Exchange A(m) and A(i)
            A(i) = Temp
        Next i
    End If
End Sub


Private Sub Command1_Click()
    Dim strArray(9) As String * 1
    Dim x As Long
    
    'create array of values 1..9
    For x = 1 To 9
        strArray(x) = Trim(CStr(x))
    Next x
    Perm strArray(), 6, 6 'Of course you could use variables For this
End Sub


Public Sub PrintPerm(A() As String * 1, n As Integer)
    Dim i As Integer
    Dim strVar As String

    For i = 1 To n
        strVar = strVar & (A(i))
    Next i

Dim combspath
combspath = "C:\permutations.txt"
Dim filenum As Integer
filenum = FreeFile
Open combspath For Append As #filenum
Print #filenum, strVar
Close #filenum
Text1.Text = strVar
End Sub

It all works fine, but I don't really understand it, and now I need just numbers 1-6 combinations.

Can anyone tell me how to modify so it does that?

Thanks

[Starman]

Looks like you've already done that.
In the line:

Perm strArray(), 6, 6 'Of course you could use variables For this

I imagine the 6's were 9's when you got it ?

[Jez1]

Hmm. That's what I thought, but it doesn't seem to produce anywhere near enough permutations -
I'm no mathamatician but I guess there's more than 750, which is what this gives me.

[Starman]

720

number 1 can appear in any of the 6 places
number 2 can appear in any of the other 5 places
number 3 can appear in any of the other 4 places
etc.

so total number is 6x5x4x3x2x1 = 720

[Jez1]

So there really is only 750 permutations of the numbers 1-6?

Wow, thanks.

[Kalkewl8ter]

There are 6! permutations, because of what Starman said:

"number 1 can appear in any of the 6 places
number 2 can appear in any of the other 5 places
number 3 can appear in any of the other 4 places
etc.

so total number is 6x5x4x3x2x1 = 720"

If you have a permutation of n distinct objects (doesn't have to be numbers), it would be n! or n*(n-1)*(n-2)*(n-3)*...*3*2*1

[A$$Bandit]

While we're talking about permutations, I think you should all go out and buy Amon Tobin - Permutation. A VERY classy album indeed, experimental drum 'n' bass at its best.