Find the sum of all 4-digit numbers formed by using digits $0, 2, 3, 5$ - possible formula for competitive exam 
Find the sum of all 4-digit numbers formed by using digits 0, 2, 3, 5 without repetition

There is a similar question in this site and Eric Tressler has provided a clear method to solve such questions. I have solved this question usingthe same approach and answer I got is $64440$
Today, came across a general formula in careerbless.(attaching it here)

According to this formula, I could solve this question as

$(4-1)!(0+2+3+5)(1111) - (4-2)!(0+2+3+5)(111)\\
 =3!×10×1111-2!×10×111\\=64440 $

Will this can formula be true in all cases where we find sum of all the n digit numbers formed by using n digit digits in which one digit is zero?
 A: Yes, it will work, but I would not suggest memorizing it.  It is good to understand where it comes from.  Let us start with the case that you have $n$ different digits, none of which are zero, and ask for the sum of all the $n$ digit numbers you can form.  There are $n!$ numbers, one for each order of the digits.  A specific digit $a$ appears in each position $(n-1)!$ times, so if we sum up its contribution we get $a \times (n-1)! \times (111\dots n \text{ times})$  Then summing over the digits gives the first term in your formula.  Then if one of the digits is $0$, we need to account for the fact that we do not consider numbers starting with $0$ to be $n$ digit numbers.  The sum of all the numbers starting with $0$ is the sum of the $n-1$ digit numbers formed from the remaining $n-1$ digits.  We use the same formula as before, but decrease $n$ by $1$ and get the subtraction term.
A: If you do want to use a formula, it is better to put it in a self-explanatory form.
$n$ distinct digits form a $[n\;\; columns\times n!\;\;rows]$ phalanx,
and each column repeats each digit $\dfrac{n!}{n}$ times,
thus $\;\;\dfrac{n!}{n}\times\left[\text{ sum of digits}\right]\times[10^0 + 10^1 + ...10^{n-1}]$
Subtraction for numbers with a leading zero should be obvious 
