Happy to help. The game requires us to assign six teaching assistants - R S T V Y Z - to three courses - L M P. Each assistant is assigned to exactly one course and each course will have at least one assistant assigned to it. This rule tells us that each course must have at least one assistant but could have as many as 4 since we have to assign 6 assistants to three courses, resulting in the following potential ratios of assistants to courses:
4-1-1 3-2-1 2-2-2
The following rules apply:
(1) M must have exactly two assistants. This rule tells us that the scenario 1 above is impossible, so we are down to two options:
(a) 3-2-1 (b) 2-2-2
The question asks us if no other student is assigned to the same course as Ramos. This means we have exactly one course with only one TA assigned to it - Ramos, meaning we are in scenario A above.
Since we only have two courses to assign 5 assistants - S T V Y Z, and we know that S&T must be assigned to the same course, and V& Y must be assigned to different courses, we can infer that S&T are not assigned to M because it would force V&Y to be assigned to the remaining course along with Z. We also cannot assign neither Y nor Z to P because it would result in S&T being assigned to M, thus we can conclude that S&T must be assigned to either L along with Y or V.
L: S T /Y /V M: /V /Y Z P: R
or to P along with V:
L: R M Y Z P : S T V
We can see that in both possible scenarios Z must be assigned to M - (D).