There are 6 jobs with distinct difficulty levels, and 3 computers with distinct processing speeds. Each job is assigned to a computer such that: - The fastest computer gets the toughest job and the slowest computer gets the easiest job. - Every computer gets at least one job. The number of ways in which this can be done is _________.

Let the 3 distinct computers be:

• $C_1$: fastest computer

• $C_2$: medium-speed computer

• $C_3$: slowest computer

Let the 6 distinct jobs ordered by difficulty be $J_1, J_2, J_3, J_4, J_5, J_6$, where $J_1$ is the toughest and $J_6$ is the easiest.

1. Fixed Assignments:

   • $J_1$ must be assigned to $C_1$.

   • $J_6$ must be assigned to $C_3$.

   • Because of this, both $C_1$ and $C_3$ already have at least one job.

2. Remaining Jobs:

   • There are $4$ remaining jobs: $\{J_2, J_3, J_4, J_5\}$.

   • Each of these $4$ jobs can be independently assigned to any of the $3$ computers ($C_1, C_2, C_3$).

   • Total unrestricted assignments for the $4$ jobs = $3^4 = 81$.

3. Condition that every computer gets at least one job:

   • Since $C_1$ and $C_3$ already have at least one job, we only need to ensure that $C_2$ gets at least one job.

   • The invalid cases are those where $C_2$ gets $0$ jobs (i.e., all $4$ jobs are assigned only to $C_1$ or $C_3$).

   • Number of such invalid assignments = $2^4 = 16$.

4. Total Valid Assignments:

   $$\text{Number of ways} = 3^4 - 2^4 = 81 - 16 = 65$$