Summary

Here is the strategy for working through grouping games:

  • Read through the premise of the game and identify the task you are asked to solve.
  • Create the roster of elements you are asked to place into groups.
  • You may also want to create a second list of the groups into which you will be dividing the elements.
  • Symbolize the conditions.
  • Create a diagram that will best represent the multiple groups into which we will arrange the elements of the game. Within each group, represent the number of elements that can be possibly placed in that group.
  • Like the strategy we use for other games, write the conditions for the game to the side of the diagram.
  • Look at the conditions that are written at the side, and see which ones can be transferred into the diagram. Always start with the conditions that are fixed.
  • Once you have transferred the information from the conditions to the diagram, look over the conditions again to see if any new information can be deduced when considered together with your diagram.
  • Read the question.
  • Determine what additional information (if any) is provided by the question itself.
  • Add that new information provided by the question to the diagram, and again review the conditions on the right of the diagram and see if you can make any more deductions about elements that can be put into a group on the diagram.
  • Solve the question by consulting your diagram. If necessary, check each possibility against your diagram.
  • When you proceed to the next question, remember that any information that was given by the previous question must not be carried over to the next question.

Sample Grouping Game One

Ten freshmen students, L, M, N, O, P, Q, R, S, T, and U, arrive at Egghead University for their first term of college. They are told that they have been assigned to one of three dormitories: Genius Hall, Intellectual Center, or Brainy Building. At least three of the ten students are assigned to each dormitory. M, P, and U are all assigned to separate dormitories. L, N, and T are all assigned to the same dormitory. Both O and R are assigned to Intellectual Center.

Next LSAT: January 26

Question Setup Walkthrough

Now that we have added the new information provided by the question to our diagram, let’s review the conditions we have on the right of the diagram and see if we can deduce anything further about how to place the students in the groups. Again, let’s look at the requirement that L, N, and P are in the group of four students. Is there room in the Genius dorm for four students? No. Therefore, we must place the set of L, N, and T into the Brainy dorm and also move into that dorm the fourth space, which would be held by U or P.

First, what are we being asked to do here? Out of a group of ten students, we need to separate them into three groups based on the dormitory in which they will live. Once we have read through the premise of the game, realized it is a grouping game, and identified the task we are asked to solve, the next step in our game strategy is to create the roster. In this case, the roster of elements is the list of students.

Roster: L, M, N, O, P, Q, R, S, T, and U

For grouping games, you may want to create a second list of the groups into which you will be dividing the elements. In this case, the groups are the three buildings: Genius Hall, Intellectual Center, or Brainy Building.

Groups: Genius, Intellectual, Brainy (g, i, and b)

Notice that we represent the group names with lower case letters, rather than the capital letters we used to represent the students. It’s always a good idea to make your abbreviations for the roster of elements and the abbreviations for the groups look different so you don’t run the risk of mixing them up during the process of solving the game.

Okay, according to our game-solving strategy, our next step is to symbolize the conditions. In this case, one of the conditions is actually contained within the premise of the game: we are told that at least three students are assigned to each of the three dormitories. What additional information can we deduce from that statement? Well, if there are ten students, and at least three will be in each dorm, that would give us three students in three dorms, for a total of nine students. Since we have ten students, not nine, we can conclude that one of the dormitories will have four students, and the other two will have three students each. We don’t know yet which of the three dormitories has four students, and which two have three, but for now, we can summarize this simply as 3-3-4.

The next condition (the first in the list of conditions following the premise) tells us that M, P, and U are all assigned to separate dormitories. This means that if M is in a dorm, then neither P nor U can be in that same dorm. (This applies to the others as well, that is, if P is in a dorm, then M and U are not in that dorm, etc.) We can use the same symbol we used earlier to indicate “not,” the ^ symbol, and represent this condition as:

  • If M ^P, if P ^M
  • If M ^U, if U ^M
  • If P ^U, if U ^P.

As you can see, that one sentence condition has given us a lot of information. We know that these three students are assigned to the three different dormitories. How else could we represent this condition symbolically? Rather than writing three lines, a shorter way might be to just use the ? symbol. In this case we can use it to represent the fact that if one of the students is assigned to a dorm, the others cannot also be assigned to that dorm:

M ? U ? P

The next condition tells us that students L, N, and T are all assigned to the same dormitory. Using the opposite symbol from that which we used for the previous condition, we can represent this as:

L = N = T

Again, what other information can we deduce from this condition? Since we know that two of the dormitories will have three students and one will have four, we know that either only these three students will live in one dormitory, or one additional student could also be assigned to the same dormitory and that this dormitory would then be the one that has four students. Now, let’s think about the previous condition. We know that students M, U, and P all live in separate dorms (M ? U ? P). That means that one of them, though we don’t yet know which one, will be assigned to the same dorm as L, N, and T. Therefore, we can modify our condition to represent this new information:

4 = L, N, T and (M, U, or P)

The third condition tells us that both O and R are assigned to Intellectual Center. We can represent this condition as:

i = O and R

Now, let’s create a diagram to help us think about the problem. What kind of diagram will best represent the arrangement of elements in the game? Unlike ordering games or network games, we don’t need a complicated diagram. We only need to represent our three groups and we can just use three boxes for this. (Note: in this diagram the groups are labeled with their full names for clarity, but on the actual exam, you will probably want to use the abbreviations in order to save time.) Within each group box, we will put three horizontal lines for the three elements we will place in those groups. (Remember that one of the three dorms has four students. Since we don’t know yet which one, we can represent that underneath each of the three group boxes with a horizontal line and a question mark.) Like our other games, we will write the conditions for the game to the side.

If we look at the second to last condition we have written to the side, 4 = L, N, T and (M, U, or P), we remember that L, N, T and one other student are in the dorm that has four students. Because we already have O and R in the Intellectual dorm, there is not enough space left in that dorm. Therefore, we know that the Intellectual dorm does not have four people in it, so we can eliminate the line from underneath this group. Further, if we look at our second condition which states that M, U, and P must be in separate groups, we know that one of these students must be the third one in the Intellectual group. We can represent this possibility as M / U/ P on that third line, symbolizing that one of these must be in this group.

The next step is to look over the conditions again to see if any new information can be deduced when considered together with your diagram. We know that L, N, T and either M, U, and P make up the dorm that has four of the students, but we don’t know yet if these four are in the Genius or Brainy dorms. It seems as though this is all the information we can fill in so far.

Question One

If M and Q are assigned to the same dorm, which of the following cannot be true?

(A) O and U are in the same dorm.
(B) R and P are in the same dorm.
(C) Q and S are in the same dorm.
(D) U and L are in the same dorm.
(E) M and T are in the same dorm.

After reading the question, the first step is to determine what additional information is provided by the question itself. In this case, we are told that M and Q are assigned to the same dorm. If we look at our diagram, M and Q could fit into either the Genius or the Brainy dorms. If we look at the answer choices, we see that we are not going to be asked which students are in specific dorms, only which students are assigned to the same dorm. So, let’s arbitrarily insert M and Q into the Genius dorm to start. Once we have placed M into the Genius dorm, then we can remove it from being a possibility in the Intellectual dorm. We can also add the possibility of U or P into the Brainy dorm as well.

Now that we have added the new information provided by the question to our diagram, let’s review the conditions we have on the right of the diagram and see if we can deduce anything further about how to place the students in the groups. Again, let’s look at the requirement that L, N, and P are in the group of four students. Is there room in the Genius dorm for four students? No. Therefore, we must place the set of L, N, and T into the Brainy dorm and also move into that dorm the fourth space, which would be held by U or P.

Now that we have our diagram, let’s look at the conditions we have written at the side and see which ones can be transferred into the diagram. As before, let’s start with the conditions that are fixed. Starting with the last condition, we see that O and R will be in group i (Intellectual), so let us write O and R in two of the lines in the Intellectual group.

With this additional information on our diagram, let’s now look at the answer choices. Remember that we are asked to figure out which of the following choices cannot be true. For choice A, can O and U be in the same dorm? Looking at our diagram, we see that O is in Intellectual. Can U be in this group? Yes, we have it listed as a possibility for the third spot. For choice B, can R and P be in the same dorm? Again, R is in the Intellectual group, and P is listed as a possibility. For choice C, can Q and S be in the same dorm? Well, we’ve placed Q into the Genius dorm. Can S fill that empty spot? If we look through the conditions, we don’t see any restrictions on where S can be. So that too seems like a possibility. (And remember, we are looking for the choice that is not possible.) For choice D, can U and L be in the same dorm? U is listed as a possibility for the Brainy dorm, which is where we have placed L, so this too is a possible arrangement. For choice E, can M and T be in the same dorm? We have placed M in the Genius, and T in Brainy. This is not a possibility, so choice E is the correct answer. (Note: we arbitrarily placed M and Q into the Genius dorm. In this case, it doesn’t matter because both the Genius dorm and the Brainy dorm have the same set-up, that is, they can either have three or four students. If you are not sure about this, work through the problem again with M and Q in the Brainy dorm instead.)

Let’s try a second question for the same game. What do we need to remember before utilizing our same diagram to answer the next question? Any information that was given by the previous question must not be carried over to the next question. Only information from the original premise and conditions can be applied to each question.

Question Two

Which of the following must be true for the group of ten students?

(A) Exactly three students are assigned to Genius Hall.
(B) Exactly three students are assigned to Intellectual Center.
(C) Exactly three students are assigned to Brainy Building.
(D) Exactly four students are assigned to Genius Hall.
(E) Exactly four students are assigned to Intellectual Center.

We know that two of the dormitories will have three students and one will have four. Do we know anything about which of the three dormitories could have four and which must have only three? Let’s look at our diagram again. (Remember, we need to look at our original diagram, not the one that we made for the last question that contained additional information.)

We are asked about how many students can be assigned to each dorm. As we’ve drawn it, the group of four students could be in either Genius Hall or the Brainy Building, not in Intellectual Center. Choice B, that exactly three students are assigned to Intellectual Center, is the correct choice. Choices A, C, and D are all possible arrangements that satisfy our conditions, but we are asked which choice must be correct, not what is merely possible. Choice E is not correct, since we know that only three students are assigned to Intellectual Center.

Let’s try one more question for this game.

Question Three

If U is assigned to the same dorm as S, how many distinct groups of students could be assigned to Genius Hall?

(A) 1
(B) 2
(C) 3
(D) 4
(E) 5

These particular types of questions are quite common in the games section and can be time-consuming because we are asked to figure out how many possibilities will satisfy a given arrangement. Let’s start with what we know. In the question, we are told that U and S are in the same dorm. Now, we know that they can’t be in Intellectual Center since we don’t have enough spaces available to fit two students in — only space enough for one additional one besides O and R. Therefore, U and S must be in either Genius Hall or in Brainy Building. Since we don’t know which, we need to draw out both possibilities. We will label these possibilities A and B for locating them in Genius Hall or Brainy Building.

Let’s look at our conditions again. We know we have the block of four to fit into a group. Can the group of four students be in the same dorm as U and S? No. Let’s then put the group of four into the other building. Remember, we need to do this for both possible arrangements, A and B. When we do this, we see that we have one blank remaining, which we can fill in with the one remaining student, Q.

So, we are asked how many distinct groups of students could be assigned to Genius Hall. In possibility A, where S, Q, and U are assigned to this dorm, there is only that one set. In possibility B, which we have assigned to L, N, T and either M or P, there are two possibilities (L, N, T, and M or L, N, T, and P). Therefore, we have a total of three possible arrangements of students in Genius Hall, so the correct choice is C.

Sample Grouping Game Two

In the front window of Peter’s Pet Palace, there are three cages — cage one, cage two, and cage three — that contain a total of seven cats. Each cat is a different color: brown, black, white, gray, tabby, calico, or orange. The gray cat and the tabby cat are not in the same cage. The black cat is in cage three only if the white cat is also in cage three. Each cage contains a different number of cats. The orange cat is in cage 2. There is at least one cat in each cage.

Okay, once we have read the premise, we can see that this is a grouping game. What are the groups? What are the elements we are asked to arrange into these groups? The groups are the three cages that contain the cats. The cats are the elements we need to arrange. Unlike the previous game, the elements are not identical. Rather than ten students, we have seven cats of varying colors.

Our roster might look like this:

Roster: Br, Bl, W, G, T, C, O

Note that we had two colors starting with B, so we just used Br and Bl to help us differentiate between those two cats.

Groups: (cages) 1, 2, and 3

Next, let’s symbolize the conditions. The gray cat and the tabby cat are in different cages can be summarized as: If G, ^T or if T, ^G.

We can also write this as G ? T.

The second condition tells us that the black cat is in cage three only if the white cat is also in cage three. We can write this as: 3: if B, then W

The next condition tells us that each cage contains a different number of cats. This can be represented as 1 ? 2 ? 3

We are told that the orange cat is in cage 2. This is simply represented as: 2: O

Finally, we are told that there is at least one cat in each cage. We can write this as: 1, 2, 3 ? 0

What else does this statement tell us? Well, if there are three cages, each containing one cat, and we have seven cats all together, then the maximum possible number of cats in any one cage would be five. (Since one cat would be in each of the other cages.)

Let’s draw a diagram to help us think about the problem.

Is there any information that we can extract from the conditions and put onto the diagram? Let’s start with the fixed conditions. We can put at least one line in each cage since we know there is at least one cat in each one. We can also place the orange cat (O) in cage 2.

Can we place with certainty any information from the other conditions? We could also include the statement about the cats that might be in cage 3 (3: if B, then W). Though since we don’t know yet where any cats (other than the orange one) are, let’s wait on this one until we’ve read the question.

Question One

If cage 1 contains only the brown cat, then which of the following may be true?

(A) Cage 2 contains only the orange, black, and calico cats.
(B) Cage 3 contains only the black, white, and calico cats.
(C) Cage 2 contains only the tabby, calico, gray, and orange cats.
(D) The black and white cats are the only two cats in cage 3.
(E) The gray cat is in the same cage as the white cat.

After reading the question, the first step is to determine whether there is any additional information provided by the question. In this case, we are told that cage 1 contains only the brown cat, so we can add that to our diagram. Next we want to consider the original conditions and determine whether in combination with our new information, we can move anything else to our diagram. In this case, we know that there is a different number of cats in each cage. Since there is only one cat is cage 1, we know that there must be at least two in both cages 2 and 3. Since there are seven cats total, we also know that we need to place six cats into cages 2 and 3. It cannot be exactly three cats in each cage or that would violate the rule of there being a different number of cats in each cage. We can make a note of that at the bottom of groups 2 and 3.

Now let’s work through the answer choices, keeping in mind that we are being asked for which arrangement may be true. Choices A and B cannot be correct since they place only three cats in either cage 2 or 3, and we just concluded that it is impossible to have only three cats in these cages since that would violate the 1 ? 2 ? 3 rule. Choice C does not violate the same rule, though when we look over our conditions, we see that it violates the G ? T rule, so we can eliminate this option. Choice D puts only the black and white cats in cage 3. We know that the black cat can only be in this cage if the white cat is present, which is the other part of this answer choice. What happens if we add these two cats into our diagram in cage 3? Keep in mind that only the brown one is in cage 1, we already have the orange one in cage two, and we still need to fit the other three cats into a cage, and the only one available is two.

When we do this, we can see that this arrangement will violate the G ? T rule, so this is not a possibility. Choice E is the only one left, but let’s check it to make sure. The gray cat is in the same cage as the white one. Is this possible? Let’s try placing the two cats into cage three. We know we need to have the tabby in another cage from the gray cat, so we must put the tabby in cage two. The other cats can be in either cage as long as we can figure out an arrangement that places a different number of cats in each cage. Here is one possibility, though there are several:

Let’s try a second question for this same problem.

Question Two

If the black cat and the tabby are two of the cats in cage three, then which of the following must be false?

(A) The gray cat and the calico are the only two cats in cage one.
(B) The calico and the brown cat are also in cage three.
(C) There are two cats in cage two.
(D) Besides the black and tabby cats, there are two more cats in cage three.
(E) The orange cat and the gray cat are in the same cage.

After reading the question, let’s add the new information to our diagram. Remember, we will go back to our original diagram, not the one on which we added information specific to the last question.

We know that if the black cat is in cage three, then the white cat must also be in cage three, so we can add that information to our diagram.

Now let’s work through the answer choices with the help of our diagram, keeping in mind that we are looking for the choice that must be false. Starting with choice A, can we place the gray and calico cats in cage one? Well, if we have two cats there, we need to have five cats total in cages two and three. We already know where four of those are (the black, tabby, and white cats are in cage three, and the orange cat is in cage two), so we would just need to place the brown cat. Can we place the brown cat in cage two? No, that would put two cats in both cages one and two, violating the 1 ? 2 ? 3 rule. Can we put it in cage three? Yes, so this choice is acceptable. Choice B puts the calico and the brown cat in cage three. If we add these two cats to the three we have already placed in cage three, that would give us five cats there, leaving only two for cages 1 and 2, again violating the 1 ? 2 ? 3 rule. This choice must be false, which is what we are looking for. Let’s look at the other answer choices, though, just to convince ourselves. Choice C places two cats in cage two. That would be the orange cat plus one other. Can we still place the remaining cats and satisfy the other conditions? Yes, here is one such arrangement.

Choice D puts a total of four cats in cage four. Is this possible? Yes, the last example actually satisfies this condition, too. Choice E requires the orange and gray cats to be in the same cage, and we can easily see that this is a possible arrangement. Therefore, the one choice that must be false is choice B.

Next LSAT: January 26