Average of all 3 digit numbers with distinct digits

Question

I tried adding up all the digits in each digits place and average them, but I got some decimals.

I don't know if I am doing it correctly

Can someone please explain this to me

Answer 1

The only thing that makes this hard is that the numbers can't start with $0.$ If they were allowed to start with $0,$ it would be easy, so let's do the problem on the assumption that numbers are allowed to start with $0,$ and then adjust the result.

If numbers can start with $0$, there are $10\cdot9\cdot8=720$ nunbers with distinct digits. By symmetry, each of the digits appears the same number of times $(72)$ in each column, and since the sum of the numbers from $0$ t $9$ is $45,$ the sum of the numbers is $$72\cdot45\cdot111=359640,$$ where the $111$ comes from the fact that we are summing the ones, tens, and hundreds columns.

Now we just have to subtract out the sum of the numbers that started with $0$. There are $9\cdot8=72$ of these, since the last two digits must be distinct, and neither can be $0.$ So, to finish this off, we need to subtract the sum of the two-digit numbers with distinct non-zero digits, and divide by $648(720-72).$

I'll leave that to you.