jmc

First we place the $0$, which we only have four options for (everywhere but the first digit). Then we have 4 remaining places to put the last 4 digits, two of which are not unique (the fives), so there are $\frac{4!}{2!}$ options for arranging the other 4 digits. This gives a final answer of $\frac{4\times 4!}{2!}=\boxed{48}$.