text: what is the possibility of 5-digit even number from 0-9 with condition the 2nd number is either 3/6/9? 0-9 form 5 digit even number the 2nd number is either 3/6/9 assuming these are the only restrictions: { 3/10 possibility (3, 6, or 9) for the 2nd digit, 5/10 possibility (0,2,4,6,8)  for the 5th digit   } there are 3*5=15 combinations of those successful outcomes, which is 15% of the number of all possible combinations (00-99).  since there are as many possible combinations of the other digits  (1000) for successful outcomes as for the general population (also 1000), those digits are irrelevant to determining the proportion of matches to the general population, so leading zeroes don't matter and neither do a 4 and a 2 in 3rd and 4th digits, as long as the overall population shares the restriction.  it's still 15%.  however, the manner in which the number is chosen could be relevant for describing the probability; for example, if the number is chosen one digit at a time such that no digit can repeat itself (like a lotto), its probability of matching the desired outcome would be different.  ===  if the value of each digit can only be used once, the overall population size is only 10!/(10-5)! = 10!/5! = 10*9*8*7*6 = 30240. if there are restrictions like "no leading zeroes", then multiply by 9/10, but unless there are other rules, matches are still distributed evenly across the other 3 digits.  as long as the order digits are selected does not matter 3/10 for digit2 5/10 for digit5 (8*7*6)/10 for the remaining digits1,3,4 (start at 8, since two are already taken in order to guarantee the match)  so (3*5*8*7*6)/(10!/5!)  =~ 16.67%  if leading zero is not allowed, then distribution is [ (8*3*7*6*1) (where digit5 is 0) +(7*3*7*6*4) (where digit5 is {2,4,6,8}) ] / (9*9*8*7*6)  ==  (8+28) / 216 = ~16.67%   if each digit is selected in order and may not be re-used, then it's more complicated.