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(The complete setup for this game can be found here: lsat/viewtopic.php?t=8562)
The correct answer choice is (C)
This is another potentially time-consuming Global, Cannot Be True question. Questions of this type can sometimes be solved by referring to Not Laws; however, we would not expect Question #23 to be this simple. And indeed, the only way to proceed is by the process of elimination: any answer choice that could be true will be incorrect. Proving that a given answer choice is possible, however, need not involve the creation of a brand new diagram. For several of the answer choices listed below, local diagrams already exist from earlier questions. These local diagrams provide a shortcut that can quickly prove that a given answer choice could be true, and is therefore incorrect.
Answer choice (A) is incorrect. If L and R are the only kinds of flowers in bouquet 1, then neither of them can be in bouquet 3 (first rule). Bouquet 3 must have at least one flower other than S, and the only remaining flowers are T and P. We cannot have T without P (contrapositive of the last rule), which is why bouquet 3 must contain P:
Note that this solution is identical to the solution to Question #20. Answer choice (A) can be eliminated on that basis alone, as it represents a viable solution to the game.
Answer choice (B) is incorrect as well. If P and T are the only kinds of flowers in bouquet 1, then neither of them can be in bouquet 3 (first rule). Bouquet 3 must have at least one flower other than S, and that flower can be none other than R. Bouquet 2, in turn, must have at least R and S (but can also contain other flowers as well):
This solution does not violate any of the rules in the game, which is why answer choice (B) could be true and is incorrect.
Answer choice (C) is the correct answer choice. If L, P and R are the only kinds of flowers in bouquet 2, and bouquet 3 must have exactly two flowers in common with bouquet 2, then these flowers must be none other than P and R, since bouquet 3 does not contain L:
So far, this appears to be a viable solution. However, recall that bouquet 1 cannot share any flowers with bouquet 3. The only flowers left for bouquet 1, therefore, are L and T, neither of which works. This is because L requires R (fourth rule), whereas T requires P (fifth rule). Satisfying either rule would require bouquets 1 and 3 to have at least one flower in common, in violation of the first rule:
As a result, answer choice (C) is the correct answer choice to this Cannot Be True question.
Answer choice (D) is incorrect, because bouquet 2 could contain P, R, and S, and no other flowers, without violating any of the rules. In this case, bouquet 3 would also contain either P or R (but not both):
Note that this solution is identical to the setup for answer choice (C) in Question #22. If you were able to reference your earlier diagram, eliminating answer choice (D) in Question #23 should take no longer than a few seconds.
Answer choice (E) is incorrect, because bouquet 3 can contain P, S, and T, and no other flowers, without violating any of the rules governing the distribution of flowers to the remaining two bouquets. In this scenario, bouquet 2 could have any given number of flowers, as long as it has P (without which bouquets 2 and 3 cannot have exactly two flowers in common) and exactly one of S or T: