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Objection’s LSAT Tips – “In/Out” GamesQuestion Type: In/Out Games Introduction In/Out games are grouping games with two groups: an in group and an out group. The rules, which rely heavily on conditional reasoning, will determine which variables are in and which are out depending on the specific circumstances. There is no ordering required. For example, a store sells six things: F, G, H, J, K, L. If F is sold, H and J must be sold. Method Diagramming conditionals is the most important skill for in/out games. You must be comfortable with contrapositives, and you must be comfortable finding inferences. In/Out games are all about making these connections between rules. In the first example below, four rules lead to more than twelve inferences. Let’s flesh those ideas out by doing some examples. Examples A bookstore carries at least one kind of the following kinds of books: J, K, L, M, N, and O. The store does not carry any other kind of books. The selection of books the store carries is consistent with the following conditions: If the store carries K, then it does not carry M. Let’s diagram these rules and see how many inferences we can find. Just a note: It’s usually unwise to spend too much time trying to list all of the possible inferences; get as many as you can and move on. For this reason, I will probably miss some inferences. But any inferences I missed were so minor, they had no affect on my ability to do the game. Here are the twelve rules which constitute my diagram (some are original rules and the rest are inferences):
Diagram rules (corresponding to above diagram and henceforth referred to as DR#): DR1. The first rule. Let’s get to the questions. 1. Which one of the following could be a complete and accurate list of the kinds of books the store carries? A) K, L, M, O How to Solve: Process of Elimination A – Incorrect. DR1 states that if K is carried, then P cannot be. 2. Which one of the following CANNOT be a complete and accurate list of the kinds of books the store carries? A) J, K, L, M, O How to Solve: Find a Rules Violation Process of elimination would take too long. This is essentially a “must be false” question, and as I have said before, locating the “must” is easier than eliminating the “could’s”. The best place to start looking for rules violations is the answer choice that brings in the most variables. Answer choice A brings in five of the six possible variables, and it also introduces a conflict. DR1 says if K is carried, M cannot be. 3. If the condition that if the store does not carry N then it does carry K is suspended, and all other conditions remain in effect, then which one of the following CANNOT be a complete and accurate list of the kinds of books the store carries? A) J, L, M, O How to Solve: Rework Diagram I hate questions like this. I spent all that time using the same diagram, and now they want me to change it? Give me a break. Thankfully, the rule they ask you to throw away usually doesn’t alter your diagram too much. In this case, you must look for a “must be false” answer. Crossreference each answer quickly with your diagrammed rules, ignoring DR3+4+9+12 as those are the rules nullified by the new condition. Doing that, you’ll see that the correct answer is E.O brings in at least one of J and N (DR7). 4. If the store sells no O, then which one of the following must be true? A) The store carries neither L nor K. How to Solve: Locate the “Must” Using Inferences Again, look for the “must”. If O isn’t sold, then L cannot be sold either (DR6). That leaves four potential books remaining: K, N, M, J. K and M cannot be together (DR1+2). That means your options are K/M, N, J. There are no conflicts between N and J, so both can be sold, as well as one of K and M. That means that, at most, three other books are sold. This is answer choice E. 5. Which one of the following could be the only kind of book the store carries? A) O How to Solve: Rules Scan The best way to solve this is to scan for the rules not involved in bringing anything else in. A quick glance at our diagram eliminates L (DR5), M (DR12), and O (DR7). This leaves J and N (choices C and D, respectively). Can J be by itself? J alone means there is nothing else. However, DR4 states that if there is no K, there must be N. J cannot be alone. This leaves answer choice D. 6. If the store sells O, then which one of the following must be false? A) The store sells M but not N. How to Solve: Locate the “Must” Using Inferences Just cross referencing our list of rules with the answer choices shows us that answer choice Amust be false: if M is sold, N must be sold (DR12). Sorry to be so short, but that is the beauty of inferences. This is actually true regardless of whether the store sells O. Sometimes an irrelevant rule is introduced in the question.  From among three law school deans – A, B, C – three business school deans – N, O, P – and three medical school deans – X, Y, and Z –, a group of five college deans will be formed. Selection is constrained by the following rules: The group must include at least one dean of each school type. This is essentially a “more” in/out game since we have nine variables for five slots. Because of the limitations set by the first rule, there are five possible distributions, as you see in my diagram: Law: A B C LawBusMed 311 1) A > ~N Explanation of diagram rules: [One of the distributions above is impossible because of a rule conflict. Try and spot it. Answer at the end.] 1. If four of the deans selected are A, O, Y, and Z, which one of the following must be the fifth dean selected? A) B How to Solve: Use Distributions to Aid in Process of Elimination This is one of the rare “must” questions that is best solved by process of elimination. Four of the five answers must be false. If one dean must fill the fifth spot, the other four cannot. Y and Z are both med schools deans. That means we are looking at either a 122 distribution or a 113. A – Incorrect. We need a 122 or a 113 distribution. We already have the one law school dean, so we cannot have another. This rules out both B and C. 2. If P is the only business school dean selected for the group, which one of the following must be true? A) A and B are both selected. How to Solve: Use Distributions to Locate the “Must” If P is the only business school dean selected, that means we are looking at a 311 or a 113 distribution. Look for rules involving P. P brings in X and Z – both med school deans. That means the distribution is 113. (You cannot have a 212 distribution because more than one law school dean precludes the possibility of more than one medical school dean.) Therefore, the answer is E, which says that XYZ (the three medical school deans) must be selected. 3. Which one of the following is an acceptable selection of deans for the group? A) A, B, N, X, Y How to Solve: Process of Elimination A – Incorrect. See DR1+2. A and N cannot be together. 4. If both B and C are among the deans selected, then the group must include either A) A or else N. How to Solve: Locate the “Must” using Distributions Both B and C being selected narrows this down to either a 311 or a 221 distribution. The 311 distribution is ABC, O, X/Y/Z. P cannot be selected because it brings in multiple medical school deans (DR5+6). N cannot be selected because it conflicts with A (DR1+2). The 221 distribution is BC, NO, X/Y/Z. A cannot be selected because the two business school deans must be N and O, as P would bring in multiple med school deans (DR5+6) and A conflicts with N (DR1+2). A – Correct. Looking at our two possible distributions above, either A or N must be selected. There is otherwise no difference between them. 5. If X is the only med school dean selected, which one of the following must be true? A) If N is selected, B cannot be selected. How to Solve: Locate the Must After reading the question prompt, we know that X is selected and Z and Y are not. Since Z is not selected, P cannot be selected (DR6). We now have five variables (A, B, C, N, and O) for the four remaining slots. Only one of A and N can be selected (DR1+2), so all three of the other variables (B, C, and O) must be selected. A – Incorrect. B must be selected.  If you’ve stuck around to see the answer to the bonus question, here it is. The 131 distribution is actually impossible. Kudos to you if you caught this. P always brings in multiple medical school deans and P cannot be with N. What does this tell you? You won’t get all the inferences right away. The key is to get enough inferences and remain openminded throughout. Final Notes For an alternative approach to in/out games and a more thorough explanation of conditional reasoning, check out this lesson in conditional reasoning by a fellow TLS user. I hope you enjoyed learning about my favorite game type as much as enjoyed writing it. Hit the forums and tell me as much… 
Ken's Introduction to the LSAT Conquering the LSAT How I Scored a 180  Article #1 How I Scored a 180  Article #2 How I Scored a 180  Article #3 Retaking the LSAT Logic Fundamentals: A Lesson In Conditional Reasoning Objection’s LSAT Tips – “Describe” Questions (LR) Objection’s LSAT Tips – “Main Point” Questions (LR) Objection's LSAT Tips  "Must Be True" Questions (LR) Objection’s LSAT Tips – “Role” Questions (LR) Objection’s LSAT Tips – “Level Ordering” Games Objection’s LSAT Tips – “In/Out” Games Objection’s LSAT Tips – “More/Less” Ordering Games Objection’s LSAT Tips – Simple Ordering Games Objection’s LSAT Tips – Multiple Group Games LSAT Prep with Work and School 
