The reason why the answer given by @Zxy123 is wrong is that it has lots of redundant counting involved.

Because when suppose you distribute job-2 to C2 (Job-1 being the hardest given to C1 and job-6 being the easiest given to C3.). And then distribute rest of the 3 jobs in $3^{3}$ ways, then suppose we distribute the 3 jobs such that J3 & J4 are given to C1, and J5 to C2. Then this will be recounted when we initially allot J5 to C2 and then distribute rest of the 3 jobs in $3^{3}$ ways.

**So the point to be Noted is that:**

Whenever you select only few objects from the group to distribute among the boxes and then again decide to distribute the left over objects to the same set of boxes then there is a chance of redundant counting.