maybe you gave me an idea..let say I make a string of letters and nubmer i.e ABCD..XYW123456789, how many letters are in english and 9 numbers..for example 23 letters and 9 number it is 32.. with for next loop running 5 times , using rnd() function and mid() function, I will take the letter or number from that position...interesting idea..
I just tried this...
dim str as string
DIM x as integer
dim mm as string
mm =""
str = "ABCDEFGHIJKLMNOPQRSTWXY123456789"
FOR x = 1 TO 5
mm = mm & Mid(str;Int(Rnd()*32);1)
x= x + 1
NEXT