In & Out Grouping vs. Two or More Groups
 by
 Aug 04, 2016
 Advice on Logic Games, LSAT
 Reviewed by: Matt Riley
Suppose we had a cast of characters — let’s call them Frank, Garfield, Henrietta, Ipecac, Jeremiah, Kougar, Lambada and Mong. Grouping games on the LSAT might ask you to combine these characters in a few different ways. There are some important things to know about the different kinds of rules in a grouping game, and what they mean in different kinds of grouping games. Let’s run through them.
In our first example, let’s say that some of these people serve on a panel about national security. This involves selecting a subgroup of our cast of characters, so we’ll call it an In & Out game. Some of them are on the panel, or “in.” Others may be left “out.” It’ll look like this:
In: _
Out:
It’s helpful and important to keep track of the “out” group. Knowing that someone’s out can be just as useful as knowing that someone is in. But think about the premise behind the game. Some of them are on the panel. When the panel meets, they’ll be together. But where are the ones not on the panel? We don’t know. They’re not at the meeting. But they might just all be at home. Or at the diner. Or anywhere. This leads to an important point:
The “out” group isn’t really a group. It’s just the list of people who aren’t in.
Now suppose you’re told that if the panel includes Ipecac, it cannot include Lambada. That’s a conditional statement:
I —> not L
Did I need to tell you to take the contrapositive?
L —> not I
Look at the original rule and its contrapositive together. Start with the lefthand, sufficient side. Notice that in both cases, the rule tells you that if one of the two is in, the other is out. We can simplify this rule to mean that I and L can’t be together on the panel. They could also have just given us the same rule by saying that the two can’t be together. We’ll write it like this:
I <––> L
Think of that symbol as a double arrow, but negated. In other words, the presence of either one on the panel is sufficient to send the other packing. They can’t both be on the panel. At least one of them has to be off the panel, or “out.” Could they both be out?
Yes, they can both be out. Again, the “out” group isn’t really a group. If they’re out, they’re not necessarily together in any meaningful way. The rule was about who’s on the panel. If one of them is on the panel, the other can’t be. But if one of them isn’t on the panel, we don’t have an arrow to follow. If you’re in doubt, look at the original conditional relationship. We can’t say anything. So they can be “together” in the out group, but they’re not really together — the out group isn’t really a group, just the list of people who aren’t in.
Let’s take another rule: if the panel does not include Kougar, it must include Garfield. That’ll look like this, contrapositive included:
Not K —> G
Not G —> K
This one tells us that if either one of the two is absent from the panel, the other must be there. This, also, we can simplify: it means that at least one of them is on the panel:
G or K
But does it mean that one of them has to be out? It doesn’t. Remember, the out group isn’t really a group. They could both be on the panel.
Now let’s suppose that instead of selecting some of these fine folks for a panel, they’re all now going to the same potluck. Each of them brings either an appetizer or a main dish. That’s grouping, too. There are two groups, and no one gets to be out. No freeloaders — everyone has to bring something:
A:
M:
Now what if I told you that Frank and Henrietta don’t bring the same kind of dish? That’s another “cannot be together” relationship, and we’ll symbolize it the same way:
F <––> H
That means that they can’t be together, in either group. They can’t both bring appetizers. They can’t both bring main dishes. Since there are only two options, that means that one of the two has to be in each group.
Now let’s imaging multiple groups. Suppose that after the potluck, each of two cars carries exactly three of the people home. Notice anything funny? We had eight people. Two cars with three people each adds up to six. We don’t know what happens to the other two. Maybe they walk home. Maybe they get eaten by rabid coyotes at the potluck. Either way, they’re not in the cars. Once again, there’s an “out” group. The setup looks like this:
1: _ _ _
2: _ _ _
Out: _ _
Again, it’s important to remember that the “out” group isn’t really a group. Those two people don’t have to be together. They’re just not in one of the cars.
Let’s take a couple sample rules. Let’s say that Kougar and Mongo can’t be in the same car:
K <––> M
Could they both be out? You bet. The rule just said they couldn’t be together. The “out” group isn’t really a group.
Or what if any car that includes Lambada must include Jeremiah:
L —> J
Not J —> Not L
Clearly, if L is in car 1, so is J. If L is in car 2, you get the idea. What if L is out, though? The wording of the rule should provide a hint: “any car.” The “out” group isn’t really a car. It isn’t really even a group. If L is out, J can go anywhere. Absolutely anywhere.
In a grouping game, any time you don’t have spots to account for all the players, you should make an “out” group. But remember, those players aren’t necessarily grouped together in any meaningful way. The “out” group isn’t really a group and so the grouping relationships don’t apply the same way.
Search the Blog
Free LSAT Practice Account
Sign up for a free Blueprint LSAT account and get access to a free trial of the SelfPaced Course and a free practice LSAT with a detailed score report, mindblowing analytics, and explanatory videos.
Learn More
Popular Posts

General LSAT Advice How to Get a 180 on the LSAT

Entertainment Revisiting Elle's LSAT Journey from Legally Blonde