9. (B)
Question Type: Unconditional
This question does not give us conditional, "if", information. Instead, it asks a question that applies to every scenario of the game: which two slots is it impossible for J to occupy at the same time?
One effective approach to unconditional questions is to check prior work. In this case, I could check work from 6, 7, and 8 to see if they help me eliminate any answer choices.
From 6, we see J can present 2nd and 5th. We could eliminate any answer choices that say J could not go 2nd and 5th. But there aren’t any.
From 7, we see J can present 2nd and 4th. We can eliminate Answer (E).
From 8, we see J can present 1st and 5th. We can eliminate (C). That leaves us only (A), (B) and (D) to consider.
If we can’t see a clever approach, we now have a max of 2 answer choices to test. For example, if (A) and (B) were possible, we would be able to circle (D) without trying it out. Using prior work saves time, especially when you need to plug in.
But, it’s always helpful to remember: the LSAT asks questions for a reason. Why might J not be able to teach wherever he wants? What do I need to remember? Which rule is very limiting?
That’s right, the KK anti-chunk! Did any of the K distribution scenarios we consider earlier involve the J? Yes. When we thought about how to squeeze 3K’s into 6 spaces, we saw that if K isn’t presented 1st, J must be 1st. In that case K will go 2nd, 4th, and 6th. A 170+ test-taker will look for ways to exploit this knowledge. Therefore, a 170+ test taker will look for an answer choice that has J going in both spot 1 and in spot 2, 4, or 6 (the spots that must be occupied by K when J is 1). Answer B has J going 1st and 4th, which we know is impossible. Therefore, (B) is our answer.
Another approach to utilizing the KK anti-chunk would be to draw new diagrams next to each answer choice we consider, placing the J’s as dictated by the choice, to see whether there’s any room for the K’s to fit. The diagram next to (B) would look like the following:
J __ __ J __ __
It’s now visually clear that we could not squeeze in three K’s without having two be adjacent to one another, and we could circle (B) without considering the remaining choices.