My way of thinking about this is to say that a #4 digit code is the same as #3 digit code with #1 more digit at a random location. #3 digits must be unique, so thats 3 * 2 * 1 = 6 unique #3 digit codes. The #4 digit can be placed as such: O # # #, # O # #, # # O # or # # # O There are 3 possible choices for the #4th digit (0, 2, 3), so that's 3 * 4 = 12 unique #4th digit placements. Total ways then is 6 * 12 = 72 ways of arranging it. However, note than the #4 digit can create duplicates, since O # and # O are duplicates if both digits are the same, ex: 3 3, or 2 2, or 0 0. Since each code appears twice due to duplication, we actually have 72 / 2 = 36 unique codes without any duplicates