### Instruction: What is the average (mean) of all $5$-digit numbers that can be formed by using each of the digits $1$, $3$, $5$, $7$, and $8$ exactly once?### Response: To find the average of a set of numbers, I need to add up all the numbers and divide by how many there are.
In this case, the numbers are all $5$-digit numbers that use each of the given digits once.
How many such numbers are there?
Well, I can choose any of the $5$ digits for the first place, then any of the remaining $4$ for the second place, then any of the remaining $3$ for the third place, and so on.
So, the number of ways to arrange $5$ distinct digits is $5 \times 4 \times 3 \times 2 \times 1 = 5! = 120$.
That means there are $120$ numbers in the set, and I need to find their sum.
How can I do that?
One way is to notice that each digit appears in each place the same number of times.
For example, the digit $1$ appears in the ones place $120 \div 5 = 24$ times, since for each of the other $4$ digits, there are $4! = 24$ ways to arrange them in the remaining places.
Similarly, the digit $1$ appears in the tens place $24$ times, and so on for the hundreds, thousands, and ten thousands places.
So, the total contribution of the digit $1$ to the sum of the numbers is $24 \times (1 + 10 + 100 + 1000 + 10000) = 24 \times 11111 = 266664$.
The same reasoning applies to the other digits, so the sum of the numbers is $266664 \times (1 + 3 + 5 + 7 + 8) = 266664 \times 24 = 6399936$.