jmc

We can solve this problem easily, just by calculating how many choices there are for each of the four digits. First off, we know there are only $2$ choices for the first digit, because $0$ isn't a valid choice, or the number would a 3-digit number, which is not what we want. We have $3$ choices for the second digit, since we already used up one of the digits, and $2$ choices for the third, and finally just $1$ choices for the fourth and final one. $2\ast 3\ast 2\ast 1$ is 12, but there are 2 zeros that have been counted as different numbers, so divide by 2 to get $\boxed{6}$.