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 Administrator
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#81271
Complete Question Explanation

The correct answer choice is (C).

Answer choice (A):

Answer choice (B):

Answer choice (C): This is the correct answer choice.

Answer choice (D):

Answer choice (E):

This explanation is still in progress. Please post any questions below!
 lsatstudier
  • Posts: 49
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#31577
Hi again. Sorry for all of the questions.

I'm trying to see what I could substitute for "M must be assigned to 1" in order to have the same effect on the assignments, but I am still struggling to figure out why C is the answer. In a question like this, do I need to try every answer choice? I drew the inference that if I have a limited amount of variables available for a particular committee, I can easily fill in who needs to be present in each committee. But why would answer A not be a possible solution?

Thank you again!
 Kristina Moen
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#31586
Hi lsatstudier,

I will assume you are asking about #23. We changed the title to reflect that.

What you want is something that will lead you to the rule that M must be on exactly one committee. All of the answer choices are statements about relative number of committees (for example, H is assigned more committees than M). One way to replace the rule that M must be on exactly one committee is to create a rule that a volunteer who is on exactly 2 committees must be assigned to more committees than M. This becomes a numerical distribution problem.

Please see LSAT Instructor Jonathan's explanation of the numerical distributions of the volunteers here:
Setup and Rule Diagrams - Game #4

There, you will find that K must be on exactly two committees and no volunteers are ever entirely omitted. So Answer Choice (C) is correct. The rule that "K must be assigned to more committees than M" means that M serves on exactly 1 committee.

Hope this helps!
 Ari
  • Posts: 22
  • Joined: Aug 27, 2020
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#79009
Hello,

I have looked at the explanation under this question, as well as the numerical distribution explanation, and I am still struggling to see why A is not a valid answer. I do understand how C is, but A creates the same conditions for M to occupy one slot. I feel like maybe my disconnect is that I don't truly understand why K being in two slots is what forces M to be in one. Can someone help? Thank you!
 Jeremy Press
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#79030
Hi Ari,

The problem with answer choice A is that it permits an extra solution that the original rule about M would not allow. It allows H to be assigned to all three committees, and M to be assigned to two committees. That arrangement would satisfy the rule in answer choice A, would not violate any of the other original conditions in the game, but would not be possible under the rule being substituted for (that "M must be assigned to exactly one committee"). So the rule in answer choice A would not have the same effect on the game, given that additional possible solution.

In answer choice C, we don't have that same possibility (of K being assigned to three committees), because the other rules (which are still in effect) stop K from being assigned to committee Z. That leaves K only two possible committees to be assigned to. If we add the rule in answer choice C, in order to satisfy that rule, we'd have to (1) put K on two committees, and (2) put M on one committee. There would be no other numerical arrangement of K and M that would work within the rules. For example, if you put K on 2 committees, but M on zero committees, then you'd have to fill the 7 remaining slots with the other three variables. You couldn't do it, because J can only be on one committee, so H and N would have to both be on all three committees (to fill the 6 remaining slots). But we already know that J and K both have to be on committee Y (so you couldn't have both H and N on committee Y). If you put K on 1 committee, and M on zero committees, you'd have to fill the 8 remaining slots with the other three variables. You couldn't do it, because J can only be on one committee, and then you'd have to fill 7 committee slots with just 2 variables (impossible, because you could only fill 6 committee slots with two variables, which wouldn't work anyway for reasons we just mentioned in the prior scenario). A bit complex, distribution-wise, but that's the reason answer choice C does have exactly the same effect as the original rule.

I hope this helps!

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