COMEDK Exam 2017 Solved Paper

There are 4!=24 numbers. Each digit occurring 3!=6 times, in the unit's, ten's, hundred's and thousand's places. We note that (2+4+6+8)=120. Thus in the over all sum there will be 120 units, 120 tens, 120 hundreds and 120 thousands.
The required sum
‌=120(1+10+102+103)
‌=120×1111=133320