I am practicing probability with, and without replacement and would like to try the following: 

From the set of numbers {0,1,2,3,...,9}, a 4-digit number is formed (where the first digit cannot be 0). To form the 4-digit number, the numbers are drawn and written down in order (if a 0 is drawn for the first digit, the 0 is put back and a number is redrawn until a number other than 0 is drawn). Find the probability that the number is divisible by 4 with, and then without replacement.

**For with replacement -** 

There are $9 \times 10^3 = 9000$ total possibilities for all 4 digits.

The number can be divided by 4 if the last 2 digits are a multiple of 4. So - $00, 04, 08,...,96$ (25 possibilities). 

So, there are 25 possibilities for the last 2 digits, 9 for the first digit, and 10 for the second digit. 

Therefore, P(divisible by 4 w/replacement) = $\frac{9 \times 10 \times 25}{9000}$

How's that look?

**For without replacement -**

Here, I am stuck! I'm not sure what it would look like? Any advice would be greatly appreciated.

Could it be something along the lines of:

First number $9C1$ (since 0 can't be picked); second number $9C1$ (since first number can't be picked), and for the last 2 digits $22C1$ since there are 22 possibilities that can be divisible by 4?