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## LSAT

### Unit 1: Lesson 4

Analytical reasoning – Video lessons

# Grouping setup questions | Video lesson

Watch one way to approach a set of questions about a grouping setup on the analytical reasoning section of the LSAT.

## Want to join the conversation?

• can someone please elaborate in the last question why Writing can not be selected. I understand the logic behind selecting Linguistics but I think Writing could be an option as well?
(1 vote) • writing cannot be the answer because it is not certain/concrete that a student must take this class. It has rules, whereas L can always be a course because it is not restricted from being combined with any other course. If the question read "what COULD be one of the three courses", then writing would be a viable option. Since this question asks what course MUST be taken, writing cannot be the answer. Hope this helps!
• on I do not understand how B is the answer.What is the difference between B and C?
(1 vote) • I need help with my videos. right from ordering setups, they can't play, they keep skipping. can you please find out the could be problem. thanks
(1 vote) • The explanation may have caused confusion.
I was thinking as follows:
if S is in, then W is out. However, the original rule in the context tells that W and P cannot be a pair.
If W is in, then S is out. However, the original rule in the context tells that S and H cannot be a pair.
It is true that L is in, but NEITHER S NOR W can be in, which leaves us only 2 courses in. This contradicts the context.

However, the rule only says if W then neither P nor S. The contrapositive logic is if either P or S, then no W. We know P is out at this point. That is being said, if S is in, then W is out. The rule only tells if W is in, then no P (and no S), but we cannot conclude W and P cannot be out in a “pair”. In fact, W and P could be a pair in out.
Same the rule says if W is in, S is out. It seems that S and H shouldn’t be in a pair. The rule says if H, then no S and no M. The contrapositive logic is if either S or M, then no H. If M is in, then H is not. We still cannot conclude that S and H cannot be out in a pair. In fact, S and H could be out in a pair.
(1 vote) • • For Question 3, could we use the knowledge that both P and W are out in combination with rule 3 to deduce the following? Since W is out and it cannot be caused by P being in, then it must be caused by S being in.

And if S is in then H is out per rule 1.

From here there are three options that have P W and H out. Which could save time from the method explained.

But it can go a step farther and get the exact answer if you look at the three remaining letters. L M and T. Looking at rule 2 shows that if T is in then M is out.

So PWHM must be the four out, thus the answer is B.
(1 vote) 