NMTC

1+2+8+8 =19 Each number appears 3 times in the 1000 100 10 and 1 place 19 x 1111 x 3 = 63327

At unit digits place ... 8 will come 6 times out of 12. 1 will come 3 times out of 12 2 will come times out of 12. So sum of all the digits at unit digits place = 8.6 + 1.3 +2.3 =57 So unit digit in the sum will be $\boxed{7}$ . And I will consider 5 as an add value for the ten's digits place numbers's sum.

And Same case will occur at the ten's digit place . i.e. 8 will come 6 times out of 12. 1 will come 3 times out of 12 2 will come times out of 12. so 57 + 5 =62 ....In the sum ten's digit will be $\boxed{2}$ . and 6 will be carried as add value..

In the next step , 57 + 6 = 63 . So in the sum hunderd's digit place number will be $\boxed{3}$ and again 6 will be carried for the add value for the next step. And at last , it will be 57 +6 = $\boxed{63}$ .

So the sum of all the given 12 numbers = $\boxed{63327}$

making "legal" multiple choice cheat: the sum is less than 9000*12 so we cut off 50%; then we use the fact that all elements in the sum are 1 mod 3 therefore the result must be divided by 3 which give us the only choice left

Using Python its really simple to do. Just wrote this script on Python 2.7

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>>> from itertools import permutations
>>> for x in permutations(str(1288), r=4):
    print x

So it prints all Possible Integers, now all I had to do was adding them.