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- ManhattanPrepLSAT1
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Atticus Finch
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Re: Q9
Answer choice (B) cannot be true because it would require our two M's, two S's, and two T's to all attend in pairs. The pairs will need to be MS, MT, and ST.
Take the position below
We've safely avoided having S attend the same session twice or on the same day. But now where can we place the final MT? It can't go on day 1 as M is already there. It can't go on day 2 as T is already there. And it can't go day 3 as T is excluded from attending a session on day 3.
No matter how you position any two of the chunks MT, MS, or ST, you will not have a session that could be attended by the final chunk without violating the constraints.
Let me know if you still have a question on this one!
Take the position below
We've safely avoided having S attend the same session twice or on the same day. But now where can we place the final MT? It can't go on day 1 as M is already there. It can't go on day 2 as T is already there. And it can't go day 3 as T is excluded from attending a session on day 3.
No matter how you position any two of the chunks MT, MS, or ST, you will not have a session that could be attended by the final chunk without violating the constraints.
Let me know if you still have a question on this one!
3 posts Page 1 of 1