jmc

First consider the two $3$s to appear in the units and tens places. Between $100$ and $500$, there are four such numbers: $133$, $233$, $333$, and $433$. Now consider the two $3$s to appear in the units and hundreds places. The numbers will be in the $300$s, so we don't need to worry about if they are between $100$ and $500$. There are $10$ choices for the tens digit, but we have already counted $333$, so such a scenario will add nine numbers. Finally, consider the two $3$s to appear in the tens and hundreds places. Again, these numbers are automatically between $100$ and $500$. There are $10$ choices for the units digit, but we again discard $333$ for a final count of nine such numbers. Thus, our answer is $4+9+9=\boxed{22}$.