If they aren't distinct, you wouldn't have 4 clean buttons, but just 3 - in which case we also know the repeating digit repeats exactly once and we get 12x3 (36

If they aren't distinct, you wouldn't have 4 clean buttons, but just 3 - in which case we also know the repeating digit repeats exactly once and we get 12x3 (36) possible combinations. With two clean buttons, it's 6 (if both repeat) + 4 (if only one repeats) = 10 and if there's 1 that's just one, and a terrible password.