jmc

For this problem, all we need to do is find the amount of valid 4-digit numbers that can be made from the digits of $2012$, since all of the valid 4-digit number will always be greater than $1000$. The best way to solve this problem is by using casework. There can be only two leading digits, namely $1$ or $2$. When the leading digit is $1$, you can make $\frac{3!}{2!1!}⟹3$ such numbers. When the leading digit is $2$, you can make $3!⟹6$ such numbers. Summing the amounts of numbers, we find that there are $\boxed{9}$ such numbers.