Find the number of ways the digit 0,1,2 and 3 can be permuted to give rise to a number greater than 2000

Please, I need explanatory answer

first digit 2 or 3, two ways

second 0 ,1,2 or 3 four ways
third Four ways
fourth Four ways
2 * 4 * 4 * 4 = 2 * 4^4 = 2 * 256 = 512
HOWEVER
If I can only use digit once in the number
First 2 or 3, say use 2 ( 2 choices)
second other ( say 3) , 0,1, (3 choices) say use 3
third 0 or 1 (2 choices)
left with one choice
so
2 * 3 * 2 * 1 = 12 ways

Did you know?

Did you know that there are only 12 possible permutations of the digits 0, 1, 2, and 3 that can be arranged to form a number greater than 2000? This limitation arises from the fact that the digit 0 cannot be the leading digit, and there are only 3 possible choices for the leading digit (1, 2, or 3). Furthermore, once the leading digit is chosen, the remaining three digits can be arranged in 3! (which is 6) different ways. So, the total number of permutations is obtained by multiplying the choices for the leading digit (3) by the arrangements of the remaining three digits (6), resulting in 18 possible permutations. However, since we need to exclude the permutations that result in numbers less than 2000 (which include 0 as the leading digit), we are left with only 12 valid permutations.