2005 AMC 10B Problem 20

Fix any one place, say the thousands place. Among all $5! = 120$ arrangements, each of the five digits sits in that place equally often (namely $4! = 24$ times), because swapping labels of the digits does not change the set of arrangements. So the average digit in that place is
$$\frac{1 + 3 + 5 + 7 + 8}{5} = \frac{24}{5} = 4.8.$$

The same holds for every place. The mean of a sum is the sum of the means, so the mean of all the numbers is
$$4.8 \cdot 10^4 + 4.8 \cdot 10^3 + 4.8 \cdot 10^2 + 4.8 \cdot 10 + 4.8 = 4.8 \cdot 11111 = 53332.8.$$

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

A number is a weighted sum of its digits, and averaging is linear, so the average number is built from the average digit in each place. Symmetry among the digits is what makes every place have the same average. This "average digit times $111\ldots1$" pattern applies to any set of numbers formed from all permutations of a digit set.

Sum everything: each digit appears $24$ times in each place, so the total is $24 \cdot 24 \cdot 11111 = 6399936$; dividing by $120$ numbers gives $53332.8$.

• Assuming the average digit is 5 (as it would be for 1, 3, 5, 7, 9) and answering 55555.
• Averaging only the smallest and largest numbers ($13578$ and $87531$), which is not the mean of all $120$ numbers.