Final answer:
The average of all possible five-digit numbers formed using the digits 1, 4, 8, 6, and 7 exactly once is 57777.2, which is calculated by determining the average contribution of each digit per place value.
Step-by-step explanation:
To calculate the average of all possible five-digit numbers that can be formed using each of the digits 1, 4, 8, 6, and 7 exactly once, we should first recognize that each digit will appear in each position (ones, tens, hundreds, thousands, and ten-thousands place) an equal number of times when forming all possible combinations of these five-digit numbers.
Since there are 5 digits, each digit will appear in each place value 4! (4 factorial) times, because for each place, we can shuffle the remaining 4 digits in 4! = 4·3·2·1 = 24 ways. So, each digit will contribute equally to the overall average when occupying a specific place value.
Now, let's calculate the average contribution for each place value:
- 1, 4, 8, 6, and 7 have an average of (1+4+8+6+7)/5 = 26/5 = 5.2 when they occupy any given place.
- Therefore, the average for each place value is 5.2 times the place value. For the ones, tens, hundreds, thousands, and ten-thousands places, this would be 5.2, 52, 520, 5200, and 52000 respectively.
The final average of all five-digit numbers will be the sum of these individual averages:
∑(average contribution for each place) = 52000 + 5200 + 520 + 52 + 5.2 = 57777.2
Therefore, the average of all possible five-digit numbers formed using the digits 1, 4, 8, 6, and 7 exactly once is 57777.2.