Question 29

Let the 3 computers be $C_1$ (fastest), $C_2$ (middle speed), and $C_3$ (slowest). Let the 6 jobs be $J_1 < J_2 < J_3 < J_4 < J_5 < J_6$ in increasing order of difficulty, so $J_6$ is the toughest job and $J_1$ is the easiest job.

The problem states:

1. The fastest computer ($C_1$) must get the toughest job ($J_6$).
2. The slowest computer ($C_3$) must get the easiest job ($J_1$).
   Therefore, the assignments $J_6 \to C_1$ and $J_1 \to C_3$ are fixed. Notice that $C_1$ and $C_3$ are each already guaranteed at least one job.

There remain 4 unassigned jobs: $J_2, J_3, J_4, J_5$. Each of these 4 jobs can be assigned to any of the 3 computers ($C_1, C_2, C_3$). The total number of unconstrained assignments for these 4 jobs is:

The second condition requires that 'Every computer gets at least one job'. Because $C_1$ already has $J_6$ and $C_3$ already has $J_1$, the condition is violated if and only if computer $C_2$ receives zero jobs from the remaining 4 jobs.

If $C_2$ receives no jobs, each of the 4 jobs must be placed on either $C_1$ or $C_3$ (2 choices per job). The number of such invalid assignments is:

Subtracting these invalid assignments gives the total number of valid ways: