Consider all nine-digit numbers consisting of one each of the digits 1 through 9.

What is the sum of these numbers?

tomarken's solution is in conformity with the various steps corresponding to my method and the said method is furnished hereunder:

Consider all possible 9 digit numbers satisfying conditions of the problem.These numbers are now arranged, WLOG, in ascending order of magnitude.

Then, the number of times any given digit i , where 1< i < 9, is the first digit in the said arrangement is (9-1)! = 8! =40320.

It can easily be observed that the said rule applies to any given digit i , where 1< i < 9, is the jth ( 1< j < 9) digit in the said arrangement is (9-1)! = 8! =40320.

Accordingly, the required sum

= ( 1+2+-------+9) * 111111111 * 40320 = 201599999798400.