How many 4-digit numbers greater than 1000 are there that use the four digits of 2012?
$\textbf{(A)}\hspace{.05in}6\qquad\textbf{(B)}\hspace{.05in}7\qquad\textbf{(C)}\hspace{.05in}8\qquad\textbf{(D)}\hspace{.05in}9\qquad\textbf{(E)}\hspace{.05in}12$
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 $\frac{3!}{2!1!} \implies 3$ such numbers. 
When the leading digit is $2$, you can make $3! \implies 6$ such numbers.
Summing the amounts of numbers, we find that there are $\boxed{\textbf{(D)}\ 9}$ such numbers.