Objection’s LSAT Tips – “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.

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.

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.
If the stand does not carry N, then it carries K.
If the stand carries L, then it carries both M and O.
If the stand carries O, then it carries J or N or both.

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.
DR2. The contrapositive of the first rule.
DR3. The second rule.
DR4. The contrapositive of the second rule.
DR5. The third rule.
DR6. The contrapositive of the third rule.
DR7. The fourth rule.
DR8. The contrapositive of the fourth rule
DR9. Combination of DR3 and DR1. If ~N -> K and K -> ~M, then ~N -> ~M.
DR10. Combination of DR5 and DR2. If L -> M+ O and M -> ~K, then L -> ~K.
DR11. Combination of DR5 and DR 7. If L -> M + O and O -> J +/or N, then L -> J +/or N.
DR12. Combination of DR2 and DR4. If M -> ~K and ~K -> N, then M -> N.

1. Which one of the following could be a complete and accurate list of the kinds of books the store carries?

A – Incorrect. DR1 states that if K is carried, then P cannot be.
B – Correct. This violates no rules.
C – Incorrect. If O is carried, then so are both P and W (DR5).
D – Incorrect. If O is carried, then so are W and P (DR5).
E – Incorrect. According to DR11, if O is carried, then so is at least one of F and T.

2. Which one of the following CANNOT be a complete and accurate list of the kinds of books the store carries?

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?

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. Cross-reference 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).

A) The store carries neither L nor K.
B) The store sells at least two kinds of books.
C) The store sells K.
D) The store carries neither L nor M.
E) The store sells at most three other kinds of books.

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.

5. Which one of the following could be the only kind of book the store carries?

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).

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.

A) The store sells M but not N.
B) The store does not sell N.
C) The store does not sell J.
D) The store sells M but not L.
E) The store does not sell M.

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.
A and N cannot both be selected.
N and P cannot both be selected.
If P is selected, both X and Z must be selected.
If more than one law school dean is selected, then at most one medical school dean is selected.

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
Bus: N O P
Med: X Y Z

Law-Bus-Med 3-1-1
2-2-1
1-1-3
1-3-1
1-2-2

1) A -> ~N
2) N -> ~A
3) N -> ~P
4) P -> ~N
5) P -> X + Z
6) ~X or ~Z -> ~P

Explanation of diagram rules:
DR1. A and N can’t be together.
DR2. A and N can’t be together.
DR3. N and P can’t be together.
DR4. N and P can’t be together.
DR5. P brings in Z and Z.
DR6. Contrapositive of DR5.

[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?

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 1-2-2 distribution or a 1-1-3.

A – Incorrect. We need a 1-2-2 or a 1-1-3 distribution. We already have the one law school dean, so we cannot have another. This rules out both B and C.
B – Incorrect. See explanation for answer choice A.
C – Incorrect. N and A cannot go together (DR1+2).
D – Incorrect. We only have space for one more dean, and P cannot be selected unless Z and X are as well (DR5+6). We would then have six deans, violating one of the cardinal rules of the game.
E – Correct. We’ve eliminated all other choices, and bringing X in by itself creates no additional issues.

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.
B) B and C are both selected.
C) C and P are both selected.
D) A, B, and C are all selected.
E) X, Y, and Z are all selected.

If P is the only business school dean selected, that means we are looking at a 3-1-1 or a 1-1-3 distribution. Look for rules involving P. P brings in X and Z – both med school deans. That means the distribution is 1-1-3. (You cannot have a 2-1-2 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
B) B, C, N, O, P
C) B, C, N, O, Z
D) C, N, P, X, Z
E) C, O, P, X, Q

A – Incorrect. See DR1+2. A and N cannot be together.
B – Incorrect. See DR3+4. N and P cannot be together.
C – Correct.
D – Incorrect. See DR3+4. N and P cannot be together.
E – Incorrect. See DR5.P cannot be there without X and Z.

4. If both B and C are among the deans selected, then the group must include either

Both B and C being selected narrows this down to either a 3-1-1 or a 2-2-1 distribution.

The 3-1-1 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 2-2-1 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.
B – Incorrect. P cannot be selected.
C – Incorrect. P cannot be selected.
D – Incorrect. P cannot be selected.
E – Incorrect. Z could also be selected.

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.
B) If O is selected, A cannot be selected.
C) If exactly one business school dean is selected, it must be N.
D) If exactly two business school deans are selected, A cannot be selected.
E) If exactly two business school deans are selected, B cannot be selected.

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.
B – Incorrect. If O is selected, A could or could not be selected. We are looking for the “must,” not the “could.”
C – Incorrect. O must be selected, so N could not be the only business school dean selected.
D – Correct. If exactly two business school deans are selected, they must be N and O (we’ve eliminated P – see above). If N is selected, A cannot be (DR1+2).
E – Incorrect. B must be selected.

If you’ve stuck around to see the answer to the bonus question, here it is. The 1-3-1 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 open-minded throughout.

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.