There are $3$ numbers that can be formed between $1000$ and $1999$, and $3\times2\times1=6$ numbers that can be formed with $2$ in the first place.  So the answer is $3+6=9$. 
How many $$4$$-digit numbers greater than $$1000$$ are there that use the four digits of $$2012$$? ($$2012$$ AMC $$8$$ Problems, Question \#$10$) 

A) $$6$$ 
B) $$7$$ 
C) $$8$$ 
D) $$9$$ 
E) $$12$$ 