Question 939219
  
There are 26 letters and 10 digits for a total of 36 character choices for each of the three positions, assuming upper- and lower-case letters are considered identical.
  
Given that the choice of characters is random, by the multiplication rule, there is a total of {{{36^3}}} possible ways to generate a 3-character code, out of which ADD is one.
  
The three letters required to spell ADD are {A,D,D}, and they can be generated in three different orders: ADD,DAD and DDA.  Thus the probability of generated theses characters is {{{3/36^3 = 1/15552}}}