Find the arithmetic mean of all the elements in a set

Question

Let Q be the set of positive rational numbers less than 1, which can be expressed with a 10-digit period in decimal representation in which all 10 digits are distinct(for example 0.12345678901234567890.....). Find the arithmetic mean of all elements in Q. I decided to sum all the numbers, which came out to be 1/2, which divided by 10! gives the answer. However I'm not sure whether the answer is right as I couldn't find a solution anywhere. Any help would be appreciated.

Answer 1

Hint: Say $x$ is one of those numbers. Say $y$ is the number obtained by replacing each digit $d_j$ in $x$ with the digit $9-d_j$. What is $x+y$?

Answer 2

Consider it digit by digit.

Common sense and symmetry tells us there were be as many numbers with $k$ as the first digit as any other. And we can easily verify that. The usual fear that a 10 digit number can't start with $0$ does not apply, etc.

If we average the values the first digit contribute we get the average value of the first digit is $\frac {\sum_{i=0}^9 i}{10}*10^{-1} = .45$.

If we do that for each digit we get the sum of the average values that the individual contributes = the average values of the numbers is $.45 + .045 +.... + 4.5\times 10^{-k} + .... = 0.49999999999..... = 0.5$.

The average is, unsurprisingly $\frac 12$.

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Another way to do it

every number is in the form $0.d_1d_2....d_9d_{10} \times 100000000010000000001000000000......$

For every $0.d_1d_2....d_9d_{10}......$ there is a corresdoing number $0.(9-d_1)(9-d_2)....(9-d_9)(9-d_{10})......$. This number is distinct from its original counter part (because $d_i = 9 - d_i$ is impossible for any $i$ much less all ten). So the sum of these two numbers is $0.999999999.......=1$.

So there are $10!$ such numbers and there are $\frac {10!}2$ such pairs. So the sum of all these numbers is $1*\frac {10!}2$.

SO the average of these numbers is $\frac {\frac {10!}2}{10!} = \frac 12$.

It actually does not matter that there are $10!$ such pairs. We don't care how many pairs there are. As all pairs add to $1$ and the average of each pair is $\frac 12$, the average of all the numbers will be the average of the averages of the pairs. And as that is constant the average is $\frac 12$.