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I've been working on the June 2007 LSAT (free on the LSAC website) and I've become stumped on one of the questions. The third logic game's final question is just confusing the hell out of me. Anyone care to explain how they get this?
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I didn't work this problem out, so I haven't checked the answer yet. I'm pretty sure it's D though because of the Guadalupe/Jamaica rule...which is read "if J, then GJ". So if you had 3 J's, then you have 3 GJ blocks (and the T in 7) making 7 total entities without the 2 Ms that must be in there. Thus, at most, you can have only 2 J's (i.e. from 1-7: GJMGJMT).
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drjay is right. You have 4 variables that must be used and 7 slots that must be filled. You also know that Martinique will be used exactly twice so your possible variable distributions are 2-2-2-1 or 3-2-1-1. 3 voyages to Jamaica would make it 3-3-1-0. IMO writing out variable distributions makes the unbalanced games a little easier. HTH
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