How many numbers have a $9$ in the left-most digit position? This must be $4!$ because the other four digits can be in any order, and each order gives you one such number.
How many numbers have a $9$ in the second digit position? The answer to this is exactly the same.
The same for the third position, and so on.
So all those nines occur $4!$ times in the $10000$ position (which contributes $4!\times9\times10000$ to the sum), $4!$ times in the $1000$ position (which contributes $4!\times9\times1000$ to the sum),
Adding those together, in total the nines contribute $4!\times9\times11111$ to the sum.
The same argument is valid for the other digits too, so the sevens contribute $4!\times7\times11111$, the fives $4!\times5\times11111$ etc.
Adding those together you get $4!\times(1+3+5+7+9)\times11111$.
If you want to generalise this to numbers with $n$ digits, you need to be able to write the number $111....111$ that has $n$ digits. These numbers are called repunits, and can be written as $(10^n-1)/9$ because $10^n-1$ is a number with $n$ nines.
So the correct general formula is $(n-1)!(a_1+...+a_n)(10^n-1)/9$.