In how many ways can the digits of $45,\!520$ be arranged to form a 5-digit number?  (Remember, numbers cannot begin with 0.)
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 $\dfrac{4!}{2!}$ options for arranging the other 4 digits.  This gives a final answer of $\dfrac{4 \times 4!}{2!} = \boxed{48}$.