2005 AMC 10B Problem 20

There are $5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 = 120$ such numbers, far too many to list. Instead, think about one place value at a time.

1. Each digit is equally common in each place. Look at the ones place. If $8$ is in the ones place, the other four digits can be arranged in $4 \cdot 3 \cdot 2 \cdot 1 = 24$ ways. The same is true for $1$, $3$, $5$, or $7$ in the ones place. So each digit appears in the ones place $24$ times. The same count works for every other place.
2. Average digit in each place. Since each digit is equally common, the average digit in any place is
$$\frac{1 + 3 + 5 + 7 + 8}{5} = \frac{24}{5} = 4.8.$$
3. Build the average number. A 5-digit number is (ten-thousands digit)$\cdot 10000$ + (thousands digit)$\cdot 1000$ + ... + (ones digit). Averaging each place separately, the average number is
$$4.8 \cdot 10000 + 4.8 \cdot 1000 + 4.8 \cdot 100 + 4.8 \cdot 10 + 4.8 = 4.8 \cdot 11111 = 53332.8.$$

The answer is $\boxed{\textbf{(C)}\ 53332.8}$.

A number is just its digits times place values, added up, and you can average each place on its own. When every arrangement of the digits is used, each digit gets each place equally often, so every place has the same average. So the answer is always "average digit times $11\ldots1$". The digit $8$ (instead of $9$) is what pulls the answer below $55555$.

Add them all up. Each digit appears $24$ times in each place, so the total of all $120$ numbers is $24 \cdot (1 + 3 + 5 + 7 + 8) \cdot 11111 = 24 \cdot 24 \cdot 11111 = 6399936$. Dividing by $120$ gives $53332.8$.