what is the sum of the three digit numbers in which there is no repetition of digits.

If the chosen digits are 1, 2 and 3: Sum of all the permutations of these three digits = (1+2+3)*2!*111
If the chosen digits are 2, 3 and 4: Sum of all the permutations of these three digits = (2+3+5)*2!*111
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If digits 1 to 9 are considered, total number of possible selections = 9C3 = 84
Each digit occurs 84 * 3/9 = 28 times
So, sum = 9*(1 + 2 + 3 + .... + 9)*2!*111 = 9 * 45 * 222 = 89910

If 0 is included, only two digits have to be selected. This can be done is 9C2 = 36 ways
Each digit occurs 8 times.
If the digits selected are 0, 1 and 2, the sum of all permutations is (0+1+2)*2!*111 - (1+2)*1!*11 = (1+2)*(222 - 11) = (1+2)*211
So, total sum of all numbers when 0 is selected = 8*(1 + 2 + 3 + ... + 9)*211 = 8 * 45 * 211 = 75960
So, required sum = 89910 - 75960 = 13950