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?