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Paul Black Paul Black is offline
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Default Calculate Last Digits

Hi peter,

They are 6 number combinations. So for 6 numbers from 49 numbers there
are 13,983,816 total combinations.
111111 could be numbers 01 02 03 04 05 06, making ALL 6 last digits
different.
321000 could be numbers 01 11 21 30 40 49.

The full list of categories are :-
111111
211110
221100
222000
311100
321000
330000
411000
420000
510000

The program will hopefully calculate the the total combinations for
ALL the categories and list them one under the other.

Thanks in Advance.
All the Best.
Paul

On Oct 5, 10:59 am, "Peter T" <peter_t@discussions wrote:
I don't follow this at all

"111111 means there are 2,887,500 combinations with all last digits
different."

Why ?

What's a combination ?
What's a category ?

321000 means there are 316,800 combinations where 3 of the last digits
are the same, 2 of the last digits are the same (but a different last
digit to the 3) and 1 last digit (but a different last digit to the 3
or 2).


I defy any one to make sense of that !

Why is 6930000 more than 2887500, seems out of sequence ?

In passing, looks like you'll be dealing with numbers over 32k. If
potentially so you should declare them 'As Long' to avoid overflows (As
Integer in VB/VBA is virtually redundant in 32bit systems). Also FWIW, no
need to disable screen updating for the sake of populating just two cells.

Regards,
Peter T

"Paul Black" wrote in message

ups.com...



Hi everyone,


I am trying to loop through ALL the combinations and count the number
of occurances of the last digit for each of the combinations.
There are 10 categories of last digit :-


111111 2887500
211110 6930000
221100 2772000
222000 105600
311100 924000
321000 316800
330000 3960
411000 39600
420000 3960
510000 396
Total = 13983816


111111 means there are 2,887,500 combinations with all last digits
different.
321000 means there are 316,800 combinations where 3 of the last digits
are the same, 2 of the last digits are the same (but a different last
digit to the 3) and 1 last digit (but a different last digit to the 3
or 2).


Here is what I have so far :-


Option Explicit
Dim A As Long, B As Long, C As Long, D As Long, E As Long, F As Long


Sub LastDigit()
Dim i As Integer
Dim LastDigit As Integer
Dim DigitCounts(0 To 9) As Integer ' This will hold counters for each
digit 0-9
Dim nDupl As Integer
Const minVal As Integer = 1 ' The minimum value in ANY
combination
Const maxVal As Integer = 49 ' The maximum value in ANY
combination


Application.ScreenUpdating = False


For i = 0 To 9
DigitCounts(i) = 0
Next i


For A = minVal To maxVal - 5
For B = A + 1 To maxVal - 4
For C = B + 1 To maxVal - 3
For D = C + 1 To maxVal - 2
For E = D + 1 To maxVal - 1
For F = E + 1 To maxVal


For i = 0 To 4
LastDigit = i - 10 * Int(i) / 10
DigitCounts(LastDigit) = DigitCounts(LastDigit) + 1
Next i


nDupl = 0


For i = 0 To 9
If DigitCounts(i) 1 Then nDupl = nDupl +
DigitCounts(i)
Next i


Next F
Next E
Next D
Next C
Next B
Next A


ActiveCell.Offset(0, 5).Value = nDupl
ActiveCell.Offset(1, 0).Select


Application.ScreenUpdating = True
End Sub


Any help will be greatly appreciated.
Thanks in Advance.
All the Best.
Paul- Hide quoted text -


- Show quoted text -