I think you have to use an inclusion-exclusion argument. The number of triples containing an e is 60. So is the number containing an f. But if you just add those two, you are double-counting the triples that contain both e and f. So you need to count how many there are of those (in fact, 24), and subtract that number. Then that gives you the answer 60 + 60 – 24 = 96 for the number containing either e or f (or both), in agreement with the previous method.