sum of all the four digit no. that can form with digits 012345 with repetition and
without repetition ....

There are four digits, i.e 1000s, 100s , 10s and units place.
Case 1: 1000s place: It can be taken by each of 1,2,3,4,5 . If one of these takes 1000s place, the other 5 digits can be arranged in the remaining 3 places in 5*4*3 = 60 ways. Hence sum of the 1000s place = (1+ 2+ 3+ 4+ 5) * 1000 * 60
case 2: 100s place: The case where 0 is chosen will not contribute to our sum. Hence it can be ignored. If one of 1,2,3,4,5 is chosen, the 1000s place can be chosen in 4 ways(excluding 0) , 10s in 4 ways, and units in 3 ways. Hence Number of ways =4 * 4* 3 = 48. Hence sum of 100s place = (1+ 2+ 3 + 4 + 5) * 100 * 48
Similarly for 10s place we get sum = (1+ 2+ 3 + 4 + 5) * 10 * 48 and sum in units place = (1+ 2+ 3 + 4 + 5) * 1 * 48
Hence total sum = 15 * 1000 * 60 + 15*111*48 = 979920.

In general if there are m digits, one of which is zero to form a n digit number, then the sum of the possibilities is
(sum of the digits)*[ $$10^{n-1} * ^{m-1}P_{n-1}$$ + $$(1111...(m-1)\ times)*(m-2)*^{m-2}P_{n-2}$$]

For the case where repetition is allowed,
Case 1 : Sum of the 1000s place =$$ (1+ 2+ 3+ 4+ 5) * 1000 * 6^3$$
Case 2: Sum of the 100s, 10s, units place = $$(1+ 2+ 3+ 4+ 5) * 111 * 5 * 6^2$$
Total sum = 15 * 1000 * 216 + 15 * 111* 180 = 3539700