AdditionPrinciple MultiplicationPrinciple AMC8 Basic
2012


Problem - 1184
How many $4$-digit numbers greater than $1000$ are there that use the four digits of $2012$?

When the thousands digit is $1$, there are $3$ possibilities because there are three choices for the digit $0$.

When the thousands digit is $2$, there are $3!$ possibilities.

Thus, the answer is $3+6=\boxed{9}$.

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