I'll help you break this one down. We have 5 technicians: R S T W X Assigned to 3 computers:
F: _ _ G: _ _ H: _ _
If you'll notice, there are 6 spots and 5 technicians. Someone is going to be included twice.
Rule 1: Sf ---> Xg Rule 2: Wf ---> Xh Rule 3: SW not together Rule 4: RT
I am going to make diagrams based on the last rule, because it limits us to 3 distinct scenarios.
1. F: R T G: _ _ H: _ _
2. F: _ _ G: R T H: _ _
3. F: _ _ G: _ _ H: R T
There are not many other inferences that I can make for scenario 1. Just remember that W and S must be separate, one in G and the other in H. 1. F: R T G: W/S _ H: W/S _
There is more that I can add to scenario 2. Computer G is full, so X cannot be there. Taking the contrapositive of Rule 1, we know that S cannot be in F. Therefore it must be in H. W must be in F, which puts X into H. Only one uncertain spot remains.
2. F: W _ G: R T H: S H
We can make similar inferences for scenario 3. There is no room for X in H. Using the contrapositive of Rule 2, we know that W is not with F. Therefore, W will be placed in G, pushing S into F. Using rule 1, we place X with G. Only one uncertain spot remains.
3. F: S _ G: W X H: R T
Using a setup like this one, you will move through the questions very quickly.
