Good days, everyone For instance, i have one number with 4 digits (2234). There are 12 times to turn this 4digits:- 2234, 2243, 2324, 2342, 2432, 2423, 3224, 3242, 3422, 4322, 4232, 4223 So these digits remain the same, but the position of digits is different. Can anyone provide solution? Thank you very much for ur kindness and assistance.
Your question is a little unclear. Not sure what you are wanting to accomplish completely. i would probably suggest putting the 12 versions of the number into an array and call it with rnd Code: MyArray(1)=2234 Myarray(2)=2243 ... ... randomize x=int(rnd(12*1)+1) debug.print MyArray(x) something like that would return your numbers randomly (kinda)
thanks for ur reply... What i meant is i was given a number (2234). So i wish to generate a program to produce 12 numbers with same digits but different position. Original number = 2234 Different position for this number :- -2234 -2243 -2324 -2342 -2432 -2423 -3224 -3242 -3422 -4322 -4232 -4223 If i insert each number into array, it looks like inconsistency.
ok, here is quick and dirty. not pretty and not dynamic but it will get the results which should be actually 16 not 12. Code: Dim x As Integer, y As Integer Dim a As Integer, b As Integer Dim c As Integer, d As Integer Dim intTemp As Integer Dim intlen As Integer intTemp = Text2.Text intlen = Len(intTemp) a = Mid$(intTemp, 1, 1) b = Mid$(intTemp, 2, 1) c = Mid$(intTemp, 3, 1) d = Mid$(intTemp, 4, 1) Text1.Text = Text1.Text & (a & b & c & d) & vbCrLf Text1.Text = Text1.Text & (a & c & b & d) & vbCrLf Text1.Text = Text1.Text & (a & d & c & b) & vbCrLf Text1.Text = Text1.Text & (a & d & b & c) & vbCrLf Text1.Text = Text1.Text & (b & a & c & d) & vbCrLf Text1.Text = Text1.Text & (b & c & a & d) & vbCrLf Text1.Text = Text1.Text & (b & d & c & a) & vbCrLf Text1.Text = Text1.Text & (b & d & a & c) & vbCrLf Text1.Text = Text1.Text & (c & b & a & d) & vbCrLf Text1.Text = Text1.Text & (c & b & d & a) & vbCrLf Text1.Text = Text1.Text & (c & d & a & b) & vbCrLf Text1.Text = Text1.Text & (c & d & c & a) & vbCrLf Text1.Text = Text1.Text & (d & b & c & a) & vbCrLf Text1.Text = Text1.Text & (d & c & b & a) & vbCrLf Text1.Text = Text1.Text & (d & a & c & b) & vbCrLf Text1.Text = Text1.Text & (d & a & b & c) & vbCrLf