If C were the case, our diagram would look as follows:
L_ _ _ _ L. In this case, we know that it is impossible for all of H to be before all of L, so according to the contrapositive of the final rule, P must not have left a message. We also know that if P did not leave a message, then neither did F (contrapositive of rule 5) and if F did not leave a message, neither did G (Contrapositive of rule 4). This would mean the only messages were H, L, and T, which is impossible as only one person (L) can leave more than one message. This would leave us with two empty spots and so it cannot be C.
For E, the problem is similar. E would give
T _ _ _ _ T, in which case we know that it is impossible for all of P to be before all of T, so F must not have left a message. And, since F couldn't not have left a message, neither could have G. Thus, we would only have P, T, H, and 2 Ls, leaving us with one empty spot.
