5 digit numbers from 1 3 3 0 0

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.

Another way to puzzle through: There are $3$ ways to arrange the digits $1,3,3$. Given one of these arrangements, say $313$ there are $3$ places to put two $0$'s:

$$3X1X3X$$

Think of each $X$ as a basket and now you have two zeros to put in the three baskets. There are $3$ ways to put both zeros in one basket. Then there are three ways to put the zeros in different baskets. So for each of the three arrangements of non-zero digits, there are $6$ ways to toss in the zeros. So $18$ is the answer.

In the simplest of terms (as a check), without using permutations and combinations: You draw from the available choices randomly to calculate all combinations.

The first number is one of 3-choices (1, 3, or 3). The second number is one of 4-choices (all that remain). The third is one of the 3-that remains. Then one of 2. And finally, the last choice.

There are 4 different ways to arrange the number 13300 (first 3 first, second 3 first, and the same with the zeros in combination.

$$3\cdot 4\cdot 3\cdot 2\cdot 1 = 72$$

Then, to account for the double $0$s and double $3$s, divide by 2 twice.

$$\frac{72}{2\cdot 2} = 18$$