How many numbers greater than 1000 can be formed using the digits 0...

To form a number greater than $1000$ using the digits $0, 1, 2, 3$ without repetition, the number must be a $4$-digit number.

The thousands place cannot be $0$, so it can be filled in $3$ ways (using $1, 2,$ or $3$).

The remaining $3$ places can be filled with the remaining $3$ digits in $3!$ ways.

Total number of such numbers $= 3 \times 3! = 3 \times 6 = 18$.

Answer: $18$