Average of numbers

Partial Solution:

The average is $\displaystyle \frac{\left(1+4+5+7+8+9\right)\times 5\times 4 \times 111}{6\times 5\times 4 }=629$ .

After a number $n$ is picked for the hundreds column, there are 5 options for the tens place and 4 options for the units place, so there are $5\times4=20$ possible combinations with $n$ as the first digit. So adding up only the hundreds digits of all of the 3 digit integers of every possible $n$ , we get $20\times100(1+4+5+7+8+9)$ We can then do the same for the tens and units places, so the total sum of all the integers together is $20\times100(1+4+5+7+8+9)+20\times10(1+4+5+7+8+9)+20\times1(1+4+5+7+8+9)=75480$ We can pick from 6 numbers for the first digit, from 5 remaining numbers for the second digit and from 4 remaining numbers for the last digit, so the total number of 3-digit integers we can form is $6\times5\times4=120$ . Therefore the average is: $\frac{75480}{120}=\boxed{629}$

Mathematica

Mean[FromDigits/@Permutations[{1,4,5,7,8,9},{3}]]

returns 629