First let me assume that the computers are $C_1,C_2,C_3$ having computing power in the increasing order as shown below:
And let me assume that the jobs are $J_1,J_2,J_3,J_4,J_5.J_6$ with increasing difficulty level as shown below:
Now as per the condition “The fastest computer gets the toughest job and the slowest computer gets the easiest job.” the situation is as follows:
Which means that $J_1$ is assigned to $C_1$ and $J_6$ is assigned to $C_3$.
Now note the second condition: “Every computer gets at least one job.”
Based on the above condition, at least one of the jobs $J_2,J_3.J_4,J_5$ must be assigned to $C_2$, or else $C_2$ computer shall remain vacant!!!
Now to calculate the number of ways in which at least one of $J_2,J_3.J_4,J_5$ is assigned to $C_2$, we find the all ways in which the above $4$ jobs can be assigned to computers and from that we subtract the cases in which no job is assigned to $C_2$.
The total number of possible ways= $3^4 =81$ [As each of the above $4$ jobs in the yellow bubble can be assigned to either computer $C_1,C_2,C_3$]
The number of ways in which in the above $4$ jobs are assigned only to $C_1$ or $C_3$=$2^4$ =16[as each of the jobs has $2$ choices]
So required number=$81-16=65$