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#### Paul Black

I have been Trying to Achieve this for Several Months.

The Code below Cycles through a Range of 6 Number Combinations ( 15

Combinations in Cells G13:L27 ) and Produces the Total 5 Number

Combinations Covered ( 90, which is Correct ) out of the 42,504 (

=COMBIN(24,5) ) Combinations that are Available with the Maximum Number

being Used in the Wheel, 24 in this Case ( the Maximum Number could be

Lower or Higher ).

How can the Code below be Adapted to ALSO Calculate ( Value to go in

Cell O17 ) how Many Combinations are Covered for 3 if 5 Please.

To Calculate the 3 if 5 Category you Need to Cycle through ALL 5 Number

Combinations that can be Constructed from the Total Numbers Used in the

Wheel ( 24 in this Case ). So if the Wheel Contains "x" Unique Numbers,

you Need to Cycle through ALL 5 Number Combinations from those "x"

Numbers. Then you Need to Scan the Wheel for Each 5 Number Combination

Produced and Compare it with Each Line in the Wheel to see if that Line

Matches the 5 Number Combination in *EXACTLY* 3 Numbers. If it does,

then that Combination of 3 if 5 is Covered and Added to the Total and

there is NO Need to Continue to Check for that Particular Combination

Any Further. You then go onto the Next Combination to Check and so on

Until ALL Combinations have been Cycled through and Checked with the

Wheel.

Basically, although there are 6 Numbers for Each Combination, I Need to

Cycle through Each 5 Number Combination that can be made from the

Maximum Number being Used in the Wheel, 24 in this Case ( 42,504

(=COMBIN(24,5) Available in Total ), and Count how Many of those are

Covered in the Wheel ( 90 in this Case ). Then of those 90 Combinations

Covered, I would like to Count how Many 3 Number Combinations ( 2,024

(=COMBIN(24,3) Available in Total ) are Covered in those Ninety 5

Number Combinations.

Code:

Sub test_for_5()

Dim a, dic As Object

Set dic = CreateObject("scripting.dictionary")

a = Range("G13").CurrentRegion.Value

For i = 1 To UBound(a, 1)

For ii = 1 To 2

For iii = ii + 1 To 3

For iv = iii + 1 To 4

For v = iv + 1 To 5

For vi = v + 1 To 6

z = a(i, ii) & "," & a(i, iii) & a(i, iv) &

a(i, v) & a(i, vi)

If Not dic.exists(z) Then

dic.Add z, Nothing

n = n + 1

End If

Next vi, v, iv, iii, ii, i

Set dic = Nothing

Range("O16") = n

End Sub

I would also like to Calculate the 4 if 5 & 2 if 5 Categories if

Possible Please.

I Hope I have Explained this Clearly Enough.

Thanks Very Much in Advance.

All the Best.

Paul