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#40236
Complete Question Explanation
(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:
PT73_Game_#4_#23_diagram 1.png
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):
PT73_Game_#4_#23_diagram 2.png
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:
PT73_Game_#4_#23_diagram 3.png
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:
PT73_Game_#4_#23_diagram 4.png
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):
PT73_Game_#4_#23_diagram 5.png
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:
PT73_Game_#4_#23_diagram 6.png
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 angelsfan0055
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#86172
is there anything in the rules that could have "hinted" (which i'm assuming the LSAT never wants to do) that L didn't need to be selected?

When I saw that B was incorrect, I was surprised because that would mean that L is not selected. This is even shown in the diagram for the answer.

When I looked at the rules, it technically doesn't explicitly say that each flower has to be chosen. Is that something I should be careful of going forward?
 Adam Tyson
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#86222
That's absolutely right, angelsfan0055! We have to pay careful attention to what the rules require, AND what they do NOT require! Never assume that every variable must be used, never assume that every group must have something in it, never assume, well, anything. Either the rules require something, or they don't! And anything that is not prohibited, either explicitly by the rules or implicitly by the inferences, must be allowed.

The "if" nature of some of these rules might be a little bit of a hint, by the way. If there is an "if" about it then it is not a sure thing!
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 Ryan Twomey
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#86245
Hey Angels fan,

A good time to pay attention to this is when the number of entities does not match the number of slots. So in a sequencing or linear game, if you have seven people and five slots, each entity might not have to get used, and each slot may not have to have an entity. That is usually my trigger to check if each entity has to be used and whether each slot needs an entity in sequencing/linear games. The example I cited was an over funded game. The same could be said in an underfunded game in sequencing/linear. If there are four entities and six slots, I would be looking at the prompt to see if each entity can go multiple times or maybe even 0 times, and if each slot can have multiple entities or maybe even 0 entities.

In grouping games, you always want to be checking for this as it is is much more common to have entities go 0 times or multiple times in a grouping game. In this case, the entities LPRST, can be used anywhere from 0-3 times according the prompt.

Additionally, the max number of flowers a bouquet could have is 4, because of the L to no S and visa versa relationship.

I sometimes write the range for each entity and each group if I think it will help me keep things straight, and if I think I can do it in under a minute. Getting that on the page tends to really help me.

Hope this helps.

Best,
Ryan

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