How many four digit numbers are there? Assume that 0 can't be a first digit. I got 9,000. Is that right?
Follow up question: How many of those four digit numbers have no repeated digits?

In order to count four digit numbers not beginning with $0$ with no repeating digits, proceed one digit at a time.
There are $9$ possibilities for the first digit.  Since the second digit must be distinct, there are $10 - 1 = 9$ possibilities.  For the third digit, we have $10 - 2 = 8$ possibilities, and for the fourth, we have $7$.
This yields
$$
9 \cdot 9 \cdot 8 \cdot 7 = 4536
$$
possible numbers.
