This is a linear game that requires us to determine the schedule of 3 short - G O P and 3 long - H N T seminars over three days. Each day exactly two seminars are given - 1 long and 1 short.
S __ __ __ G O P L __ __ __ H N T 1 2 3
The rules tell us that:
(1) T will not be given until both G & O have been given.
Since G & O are both short seminars, we can infer that it is impossible for T to be given on day 1. If T is given on day 2, it must come after G/O:
G/O G/O P T
If T is given on day 3, it must either come after G/O OR it could come either before or after P.
G/O P G/O T
P G/O G/O T
G/O G/O P <- against rule (2) T
(2) N will not be given until P has been given. This rule tells us that N cannot be the first seminar give, and P cannot be the last seminar given. It also tells us that the scenario above where T is the last seminar given on day 3 along with T is impossible because if P is on day 3, N must follow it on day 3.
If P(3) -> N (3) -> ~ T(3)
In other words, if P is given on day 3, T must be given on day 2 after G/O and N must be given on day 3 after P.
G/O G/O P T N
The question asks us if N is given some time before O, which one of the following must be true?
In other words, we have the following chain P - N - O - T because P must come before N and T must come after O and G. We can, therefore, conclude that P must come sometime before O (C).
Let's look at the other answer choices:
(A) is incorrect because N could come after G:
P G O H N T
(B) is incorrect because P could be given after G:
G P O H N T
(D) is incorrect because H could be given after O:
P O G N H T
(E) is incorrect because N could be given after H (see scenario for (B) above).
As long as P-N-O-T order is followed, it does not matter which seminars are given in between.
Let me know if this makes sense and if you have any other questions.
