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Answer
Take the numbers 1,2,3 => The three digit numbers that are formed are 123, 132, 231, 213, 312 and 321
In these cases, each digit is occurring twice in hundred's place, ten's place and unit's place => So, in hundred's place, ten's place and unit's place, each of these number must be added twice.
=> Formula is (n-1)!*(Sum of the digits)*(11...(n times)..11), where n is the number of digits.
So, for 1,2 and 3 as the 3 digits, the sum is 2*6*111.
For 1,2 and 4 as the 3 digits, the sum is 2*7*111 and so on.
So, the sum of all the numbers is 2*111*(6+7+8+8+9+10+9+10+11+12) = 180*111 = 19980
