This is a linear game that requires us to determine the order of visits to four out of six cities - H J M O S T. Each city is visited only once. We have the following rules:
(1) H and T must be included in the tour but they cannot be visited consequently.
This rule tells us that H and T must appear on our schedule, and both appear exactly once but not next to each other, there must be at least one or two other cities between them.
___ ___ ___ ___ H T
(2) if O is included in the tour, S cannot be. Conversely, we can infer that if S is included in the tour then O is not. These two visits are mutually exclusive, meaning only one of them could be visited or neither, which leaves us with two possible scenarios for a complete list of visited cities:
H T S/O J/M H T J M
O -> ~S S -> ~O
(3) If J is visited, it must be third.
J -> 3d
(4) If both J & M are included they must be visited consecutively. We can see that the only way both J and M could be included is if both S & O are excluded, meaning the complete list of cities visited in that scenario is:
H T J M
Since JM must be consecutive and H T cannot be consecutive, we can infer that H and T must be in positions 1 or 4 and MJ in positions 2 and 3 respectively (remember J must always be 3d)
H/T M J H/T
Putting all the rules together we have the following setup
(1) H/T M J H/T (2) ___ ___ __ __ H T S/O M/J
~HT (not consecutive) J = 3d
For this question, b is correct because If j is in it must be third, so we would have _ _ J M, which would violate the rule that h and t can't be next to each other.
