re:> The answer is 3^3, which is 27
27 is correct if we cannot have
[A] [A] [A]
Maybe 3^9 = 19,683 but is is way high.
Mike
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Browse All TopicsI have three boxes to have a single letter A, B, or C to be wriiten on each. How many different combinations I will end up with. The order is important: The letters can be repeated.
[A] [A] [A]
[A] [A] [B]
[A] [A] [C]
[A] [B] [C]
etc.
If you can code a routine in VB to come up with the combinations, it will be great.
Thank you.
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by: matthewspatrickPosted on 2009-10-27 at 09:56:09ID: 25674702
The answer is 3^3, which is 27.
Sub Make()
Dim arr1 As Variant, arr2 As Variant, arr3 As Variant
Dim x1 As Variant, x2 As Variant, x3 As Variant
arr1 = Array("A", "B", "C")
arr2 = Array("A", "B", "C")
arr3 = Array("A", "B", "C")
For Each x1 In arr1
For Each x2 In arr2
For Each x3 In arr3
Debug.Print x1 & "," & x2 & "," & x3
Next
Next
Next
MsgBox "Done"
End Sub