It is given that we cannot include zero in the 4 digit PIN. So, we have only nine numbers with us, namely, 1,2,3,4,5,6,7,8 and 9.
Now, there are two possible cases for this question.
Case 1: If a digit can be used only once .i.e digits can not be repeated.
Then, the number of cases are 9×8×7×6 = 3024 possible PIN numbers.
Case 2: If a digit can be used again .i.e digits can be repeated.
Then, the number of cases are 9×9×9×9 = 6561 possible PIN numbers.