I understood why my above mentioned approach is wrong.
First of understand the solution.
7 accidents occurred in a week say a1, a2,....a7.
No of different ways accident can occur is all accidents (a1......a7) have 7choices. Therefore required probability = $\dfrac{7}{7^7}\\$
The approach mentioned in my above comment is wrong because,each accident is unique(a1....a7) which can occur in any day of week.
So if apply x1+x2 ..+x7 = 7 here, 1 combination will be x1= 5, x2= 2 amd rest all equal to 0. But x1 = 5 have 7C5 combination as all accidents are unique and any 5 out of 7 accidents can go in x1. So using this approach is wrong for this problem.