jmc

We have four cases to consider:

Case 1: When there is only one digit, we have 4 choices.

Case 2: When there are two digits, we have 4 choices for the first digit and 3 choices for the second digit. So a total of $4\times 3=12$ choices.

Case 3: When there are three digits, we have 4 choices for the first digit, 3 choices for the second digit, and 2 choices for the third digit, so a total of $4\times 3\times 2=24$ choices.

Case 4: When there are four digits, we have 4 choices for the first digit, 3 choices for the second digit, 2 choices for the third digit, and 1 choice for the last digit. So a total of $4\times 3\times 2\times 1=24$ choices.

Sum the four cases up, we have a total of $4+12+24+24=\boxed{64}$ numbers.