This is a common sense question. I scanned through the answers and saw that C obviously had to be true. Of course the person who goes first will take their first choice (it’s the 3rd rule). So at least one person gets their first choice. C is CORRECT.
I read the other answers to make sure none of them jumped out, and went to the next question, sure I was right. This it the best strategy for this type of “unusual” game – LSAC is testing your ability to judge which answers are improbable and ignore them.
However, for the purposes of explanation, I made a couple of scenarios to disprove the wrong answers. Note that I did not draw these. They’re just illustrate the flow of choices. These drawings would take too long in timed conditions.
This diagram disproves A and D. J and L have their first choice, and neither P nor T have their second choice.
This diagram disproves B and E. P and T both have their second choice. And no one has their third choice (L has their first, and J has their fourth).
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Valerie says
In the second diagram, did you mean for T to choose second and P to choose third? Unless I’m mistaken, that would align with how you’ve drawn the boxes.
TutorLucas (LSAT Hacks) says
Yes, that’s right. T chooses second, P chooses third. Importantly, they both have their second choice of office, disproving answer choice (B).
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