What is the average of all possible five-digit numbers that can be formed by using each of the digits 8, 9, 1, 3 and 6 exactly once?

Are you just supposed to try out all the possibilities and count them or is there like a formula or something you can sub numbers into?

Thank you

we can build our 5 digit number using 5 different numerals

all possible numbers
= 5! or 120

In all of these, the lead digit, the 10-thousand digit, can be occupied by the "9" 24 times, the "8" 24 times, etc.
The value of those 24 9's would be
24(90000) = 24(9)10,000)
The value of those 24 8's would be
24(8)(10,000)
...
the value of those 24 1's would be
24(1)(10,000)

so the value of all the 5th digit numbers
= 24(9)(10,000) + 24(8)(10000) + .. 24(1)(10000)
= 24(10000)(9+8+6+3+1)
= 24(10,000)(27)

I could use the same argument for the thousand column
sum of all the 4th digit numbers
= 24(1000)(27)
....

same for the hundreds, the tens and the unit column

total
= 24(27)(10,000) + 24(27)(1000) + 24(27)(100) + 24(27)(10) + 24(27)(1)
= 24(27)(10,000 + 1000 + 100 + 10 + 1)
= 24(27)(11111)
= 7,199,928

To find the average of all possible five-digit numbers formed using the digits 8, 9, 1, 3, and 6 exactly once, we can use a formula rather than trying out all the possibilities.

To begin, let's analyze the situation. We know that there are 5 digits to choose from, and we want to find the average of all the possible five-digit numbers formed.

For the thousands digit, any of the five available digits can be placed in it. So, there are 5 choices for the thousands digit.

Similarly, for the hundreds digit, any of the remaining 4 available digits can be placed in it, as one digit has already been used as the thousands digit. Thus, there are 4 choices for the hundreds digit.

Continuing this pattern, there will be 3 choices for the tens digit and 2 choices for the units digit. Finally, for the one digit left, it has only one option, as there is only one digit remaining. Hence, there is one choice for the last digit.

Using the principle of counting, we can multiply the number of choices together to find the total number of possible permutations:

5 choices for the thousands digit × 4 choices for the hundreds digit × 3 choices for the tens digit × 2 choices for the units digit × 1 choice for the last digit = 5 × 4 × 3 × 2 × 1 = 120

So, there are 120 possible five-digit numbers that can be formed using the given digits.

To find the average, we need to sum up all these 120 numbers and then divide by 120.

Now, to make the calculation easier, we can see a pattern. Since the digits 8, 9, 1, 3, and 6 are all different, each digit will appear an equal number of times in each position (thousands, hundreds, tens, units, and last). So, the average digit in each position will be the average of these five numbers.

The sum of the digits 8, 9, 1, 3, and 6 is 8 + 9 + 1 + 3 + 6 = 27.
To find the average digit in each position, we need to divide this sum by 5 (the number of digits).

27 ÷ 5 = 5.4

Thus, each digit will appear as the average digit (5.4) in each position.

Finally, we can construct the average five-digit number by repeating this average digit (5.4) in each position:

Average five-digit number = 5.4 × 10000 + 5.4 × 1000 + 5.4 × 100 + 5.4 × 10 + 5.4 × 1

Simplifying this expression:

Average five-digit number = 54000 + 5400 + 540 + 54 + 5.4

Calculating this sum:

Average five-digit number = 60099.4

So, the average of all possible five-digit numbers formed using the digits 8, 9, 1, 3, and 6 exactly once is 60099.4.

By using the principle of counting and the concept of averages, we were able to find the answer without manually trying out all the possibilities.