.
It was just solved under this link
https://www.algebra.com/algebra/homework/word/misc/Miscellaneous_Word_Problems.faq.question.1123306.html
https://www.algebra.com/algebra/homework/word/misc/Miscellaneous_Word_Problems.faq.question.1123306.html
but the solution was not precisely correct.
So, I will place that solution here with my corrections.
In forming one of those 4-digit numbers, there are 6 choices for the first digit, then 5 for the second,
4 for the third, and 3 for the fourth. So the number of 4-digit numbers that can be formed is 6*5*4*3 = 360.
Imagine the list of those 360 numbers, ready to be added. Each of the 6 digits is used the same number of times;
and each of them is used the same number of times in each column. That means that in each column of the list
of the 360 numbers, each of the given digits is used 360/6 = 60 times.
The sum of the given digits is 2+4+5+6+7+8 = 32. So the sum of the digits in each column is 60*32 = 1920.
And then the sum of the 360 numbers is
1920(1000+100+10+1) = 2133120 <<<---=== My editing is in THIS LINE.
Answer. The sum is 2133120. Option C)