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

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" <(E-Mail Removed)> wrote in message
news:(E-Mail Removed)...
> 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
>

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" <paul_blac...@hotmail.com> wrote in message
>
> news:(E-Mail Removed)...
>
>
>
> > 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 -

I've narrowed it down to a few different interpretations of what you might
mean. But you win - I give up!

I'm curious now to see if someone else can understand <g>

Regards,
Peter T


"Paul Black" <(E-Mail Removed)> wrote in message
news:(E-Mail Removed)...
> 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" <paul_blac...@hotmail.com> wrote in message
> >
> > news:(E-Mail Removed)...
> >
> >
> >
> > > 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 -
>
>

Definitely strange, but I do see one pattern: All the digits add up to 6.
Subtract one from the right-most 1-digit and add one to the left-most 1-digit
to keep the total at 6, and so on. But the list is missing 600000. Other
than that I'm lost.

"Peter T" wrote:

> I've narrowed it down to a few different interpretations of what you might
> mean. But you win - I give up!
>
> I'm curious now to see if someone else can understand <g>
>
> Regards,
> Peter T
>
>
> "Paul Black" <(E-Mail Removed)> wrote in message
> news:(E-Mail Removed)...
> > 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" <paul_blac...@hotmail.com> wrote in message
> > >
> > > news:(E-Mail Removed)...
> > >
> > >
> > >
> > > > 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 -
> >
> >
>
>
>

Sorry guys,

Charlie you are quite right, I left out the 600000 category, my
appologies.
Basically, there are 13,983,816 combinations of 6 numbers. EACH 6
number combination has a last digit. The program will ideally
calculate each 6 numbers last digit category and keep a count. These
will then be listed.

Thanks in Advance.
All the Best.
Paul

On Oct 5, 6:24 pm, Charlie <Char...@discussions.microsoft.com> wrote:
> Definitely strange, but I do see one pattern: All the digits add up to 6.
> Subtract one from the right-most 1-digit and add one to the left-most 1-digit
> to keep the total at 6, and so on. But the list is missing 600000. Other
> than that I'm lost.
>
>
>
> "Peter T" wrote:
> > I've narrowed it down to a few different interpretations of what you might
> > mean. But you win - I give up!
>
> > I'm curious now to see if someone else can understand <g>
>
> > Regards,
> > Peter T
>
> > "Paul Black" <paul_blac...@hotmail.com> wrote in message
> >news:(E-Mail Removed)...
> > > 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" <paul_blac...@hotmail.com> wrote in message
>
> > > >news:(E-Mail Removed)...
>
> > > > > 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 -- Hide quoted text -
>
> - Show quoted text -

Hi. Interesting challenge from a timing point of view. It appears you
posted a solution. Are you looking to verify the solution?
I used another program and got slightly different answers. Here's what I
show:

{6}, 0}
{5, 1}, 396}
{4, 2}, 3,960}
{4, 1, 1}, 39,600}
{3, 3}, 3,963}
{3, 2, 1}, 317,082}
{3, 1, 1, 1}, 924,791}
{2, 2, 2}, 105,684}
{2, 2, 1, 1}, 2,773,289}
{2, 1, 1, 1, 1}, 6,929,822}
{1, 1, 1, 1, 1, 1}, 2,885,229}

Note that 6 has no solution. Suppose we had an ending value of 1.
Then
{1,11,21,31,41,51}
is the only possible solution However, such a sequence does not exists
since the max size is 49.
Hence, no 6's.

The only solutions I have that matches yours are:
(5,1), (4,2), and (4,1,1)

>> There are 10 categories of last digit :-

Your "Pattern" is also known as the "Integer Partitions" of the number 6, of
which there are 11.
As pointed out, you were missing "6". If you copied the solution from
somewhere else, it was probably because they left it out since there are
none.

--
HTH
Dana DeLouis


"Paul Black" <(E-Mail Removed)> wrote in message
news:(E-Mail Removed)...
> Sorry guys,
>
> Charlie you are quite right, I left out the 600000 category, my
> appologies.
> Basically, there are 13,983,816 combinations of 6 numbers. EACH 6
> number combination has a last digit. The program will ideally
> calculate each 6 numbers last digit category and keep a count. These
> will then be listed.
>
> Thanks in Advance.
> All the Best.
> Paul
>
> On Oct 5, 6:24 pm, Charlie <Char...@discussions.microsoft.com> wrote:
>> Definitely strange, but I do see one pattern: All the digits add up to
>> 6.
>> Subtract one from the right-most 1-digit and add one to the left-most
>> 1-digit
>> to keep the total at 6, and so on. But the list is missing 600000.
>> Other
>> than that I'm lost.
>>
>>
>>
>> "Peter T" wrote:
>> > I've narrowed it down to a few different interpretations of what you
>> > might
>> > mean. But you win - I give up!
>>
>> > I'm curious now to see if someone else can understand <g>
>>
>> > Regards,
>> > Peter T
>>
>> > "Paul Black" <paul_blac...@hotmail.com> wrote in message
>> >news:(E-Mail Removed)...
>> > > 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" <paul_blac...@hotmail.com> wrote in message
>>
>> > > >news:(E-Mail Removed)...
>>
>> > > > > 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 -- Hide quoted text -
>>
>> - Show quoted text -
>
>

Well, I apologize. I just attacked the problem from a different method and
got the time down to about 2 minutes.
However, I now get the same solution as you. I really don't see where the
error was in my previous code.
Are you trying to use Excel to verify the solution? I'm curious where you
got the answers.
Anyway, I'm not sure how long Excel would take to arrive at a solution, but
I'm guessing a very long time.

> LastDigit = i - 10 * Int(i) / 10

I can't follow your code to well, but in the above, the variable 'I' is
already an integer, so you are not doing much to "I"
It "appears" you meant something like "Int(I/10)*10 to extract the last
digit.
Just to mention, the last digit is also = Mod(n,10)

Your initial set are the numbers 1-49. After generating all 13,983,816
subsets, you extract the last digit of each.
That's =6*COMBIN(49,6), or 83,902,896 different Mod () operations alone.
I suggest starting with a set with just the last digit. (ie apply the Mod to
each of the 49 numbers.)
Then go into your Subset routine.

Anyway, I get the same solution as you, so please disregard my previous
attempt.
--
Dana DeLouis

<snip>

Thanks for the reply Dana,

I worked out the results using the COMBIN formula in Excel for each
category.
I am reasonably new to VBA and thought it would be a good excercise to
create the same answers using VBA. I realise now that this is a much
bigger task than I can manage.
I don't need the 13,983,816 combinations themselves, I was just trying
to produce the total combinations for each last digit category. The
code I produced was what I thought was needed to achieve this.
When you say ...
However, I now get the same solution as you. I really don't see where
the error was in my previous code.
.... does that mean you have the code to produce the results I am
looking for please?.

Thanks in Advance.
All the Best.
Paul

On Oct 10, 7:01 am, "Dana DeLouis" <ddelo...@bellsouth.net> wrote:
> Well, I apologize. I just attacked the problem from a different method and
> got the time down to about 2 minutes.
> However, I now get the same solution as you. I really don't see where the
> error was in my previous code.
> Are you trying to use Excel to verify the solution? I'm curious where you
> got the answers.
> Anyway, I'm not sure how long Excel would take to arrive at a solution, but
> I'm guessing a very long time.
>
> > LastDigit = i - 10 * Int(i) / 10
>
> I can't follow your code to well, but in the above, the variable 'I' is
> already an integer, so you are not doing much to "I"
> It "appears" you meant something like "Int(I/10)*10 to extract the last
> digit.
> Just to mention, the last digit is also = Mod(n,10)
>
> Your initial set are the numbers 1-49. After generating all 13,983,816
> subsets, you extract the last digit of each.
> That's =6*COMBIN(49,6), or 83,902,896 different Mod () operations alone.
> I suggest starting with a set with just the last digit. (ie apply the Mod to
> each of the 49 numbers.)
> Then go into your Subset routine.
>
> Anyway, I get the same solution as you, so please disregard my previous
> attempt.
> --
> Dana DeLouis
>
> <snip>

Hi. No, I used another program. I'm not sure how I would do this using
Excel vba in a reasonable amount of time.
Actually, I'd be curious to learn how you did it using Combin. I'd be
interested to learn If you can do this via formulas. I sure don't see it.
:>~

--
Thanks
Dana DeLouis


"Paul Black" <(E-Mail Removed)> wrote in message
news:(E-Mail Removed)...
> Thanks for the reply Dana,
>
> I worked out the results using the COMBIN formula in Excel for each
> category.
> I am reasonably new to VBA and thought it would be a good excercise to
> create the same answers using VBA. I realise now that this is a much
> bigger task than I can manage.
> I don't need the 13,983,816 combinations themselves, I was just trying
> to produce the total combinations for each last digit category. The
> code I produced was what I thought was needed to achieve this.
> When you say ...
> However, I now get the same solution as you. I really don't see where
> the error was in my previous code.
> ... does that mean you have the code to produce the results I am
> looking for please?.
>
> Thanks in Advance.
> All the Best.
> Paul
>
> On Oct 10, 7:01 am, "Dana DeLouis" <ddelo...@bellsouth.net> wrote:
>> Well, I apologize. I just attacked the problem from a different method
>> and
>> got the time down to about 2 minutes.
>> However, I now get the same solution as you. I really don't see where
>> the
>> error was in my previous code.
>> Are you trying to use Excel to verify the solution? I'm curious where
>> you
>> got the answers.
>> Anyway, I'm not sure how long Excel would take to arrive at a solution,
>> but
>> I'm guessing a very long time.
>>
>> > LastDigit = i - 10 * Int(i) / 10
>>
>> I can't follow your code to well, but in the above, the variable 'I' is
>> already an integer, so you are not doing much to "I"
>> It "appears" you meant something like "Int(I/10)*10 to extract the last
>> digit.
>> Just to mention, the last digit is also = Mod(n,10)
>>
>> Your initial set are the numbers 1-49. After generating all 13,983,816
>> subsets, you extract the last digit of each.
>> That's =6*COMBIN(49,6), or 83,902,896 different Mod () operations alone.
>> I suggest starting with a set with just the last digit. (ie apply the Mod
>> to
>> each of the 49 numbers.)
>> Then go into your Subset routine.
>>
>> Anyway, I get the same solution as you, so please disregard my previous
>> attempt.
>> --
>> Dana DeLouis
>>
>> <snip>
>
>