5 digit numbers from 1 3 3 0 0 How many 5 digit number positive integers can be made from the following numbers: $1,3,3,0,0$?
First I calculate all possible combinations $$\frac{5!}{2!2!}=30$$
Then I need to remove all 5 digit numbers startting with $0$ and $00$. They are not 5 digit numbers positive integers.
Starting with $00$ $$\frac{3!}{2!}=3$$
Starting with $0$ $$\frac{4!}{2!}=12$$
$12-3=9$, else I would remove numbers starting with $00$ double.
$30-3-9=18$
Is $18$ the correct answer, or where did i thought wrong?

Working from the other direction, the first digit is either $1$ or $3$, and then there are no restrictions on the remaining digits which can be permuted in $4!/2!/2!=6$ and $4!/2!=12$ ways respectively. Thus I arrive at $18$ ways, and you are correct.
