The Lineup
As the term “lineup” suggests, these games involve ordering elements in a line, from left to right or from front to back.
Before we begin, we need to study the vocabulary peculiar to these games.
Condition  Meaning 
A sits next to B.

Both conditions mean that no one sits between A and B. Although A is to the left of B in the text, that order cannot be assumed on the line. The phrase “immediately next to” is redundant; however, that style is often used on the LSAT. 
B sits immediately between A and C.

All three conditions mean that no element separates B from A, nor B from C. 
Two spaces separate A and B.

Students often confuse these two conditions. They do not mean the same thing. “A and B are two spaces apart” means that only one spot separates them.

It is essential that you master the similarities and distinctions described above. In addition to testing analytical skills, games measure your ability to notice subtle distinctions. Further, since game questions typically appear in blocks of six, misinterpreting even one condition can cost you six questions! Following is a common ordering game; one of this type—or a close variant—has occurred on every recent LSAT.
Lineup Game
There are five people—Bugsy, Nelson, Dutch, Clyde, and Gotti—in a police lineup standing in spaces numbered 1 through 6, from left to right. The following conditions apply:
There is always one empty space.
Clyde is not standing in space 1, 3, or 5.
Gotti is the third person from the left.
Bugsy is standing to the immediate left of Nelson.
Following the strategies developed earlier, we abbreviate the names by using the first letter of the name and then symbolize the conditions. “Clyde is not standing in space 1, 3, or 5" is naturally symbolized as C ≠ 1, 3, 5. “Gotti is the third person from the left” is naturally symbolized as G = 3rd.
Note: the fact that Gotti is third does not force him into space 3—he could stand in spaces 3 or 4. ”Bugsy is standing to the immediate left of Nelson” is symbolized as BN.
Our diagram will consist of six dashed lines, numbered 1 through 6 from left to right. Summarizing this information gives the following schematic:
B N D C G
C ≠ 1, 3, 5
G = 3rd
BN
1 2 3 4 5 6
Now, we decide the most effective order for placing the elements on the diagram. Following the guidelines on page 33, we look for a condition that fixes the position of an element. There is none. Next, we look for a condition that limits the position of an element. The second condition, “Gotti is the third person from the left,” limits Gotti to spaces 3 and 4. This condition, as often happens with ordering games, generates two diagrams: one with the empty space to Gotti's left and one with the empty space to his right:
Diagram I ___ ___ ___ G ___ ___
Diagram II ___ ___ G ___ ___ ___
Next, we look for a condition that connects two or more people. The last condition, BN, connects B with N. However, at this stage we cannot place it on the diagram. Finally, we look for a condition that states where a person cannot be standing. The first condition states that Clyde cannot be standing in space 1, 3, or 5. Noting this on the diagram yields
BNDCG (D “wild”)
C ≠ 1, 3, 5
G = 3rd
BN
Diagram I ~C ___ ~C G ~C ___
Diagram II ~C ___ G ___ ~C ___
Note: D is “wild” because the conditions do not refer to him. Thus D can stand in more positions than any other person.
This diagram is selfcontained. There is no need to refer to the original problem. If possible, always avoid rereading the problem. No further conditions can be derived, so we turn to the questions.
Question1
Nelson CANNOT stand in which one of the following spaces?
 2
 3
 4
 5
 6
Solution
The method of solution to this problem is rather mechanical: We merely place Nelson in one of the spaces offered. Then check whether it is possible to place the other people in the lineup without violating any initial condition. If so, then we eliminate that answerchoice. Then place Nelson in another space offered, and repeat the process.
To that end, place Nelson in space 2 in Diagram II:
~C N G ___ ~C ___
From the condition BN, we know that B must be in space 1:
B N G ___ ~C ___
Now D could stand in space 4, and C could stand in space 6—both without violating any initial condition:
B N G D X C (Where X means “empty.”)
This diagram is consistent with the initial conditions. So N could stand in space 2. This eliminates choice (A).
Next, place Nelson in space 4. Then Diagram I is violated since G is already in space 4, and Diagram II is also violated since there is no room for the condition BN:
B?
___ ___ G N ___ ___
The answer is (C).
As you read the remaining solutions, note the determining power of the condition BN.
To that end, place Nelson in space 2 in Diagram II:
~C N G ___ ~C ___
From the condition BN, we know that B must be in space 1:
B N G ___ ~C ___
Now D could stand in space 4, and C could stand in space 6—both without violating any initial condition:
B N G D X C (Where X means “empty.”)
This diagram is consistent with the initial conditions. So N could stand in space 2. This eliminates choice (A).
___ ___ G N ___ ___
Question2
Which one of the following is a possible ordering of the 5 people from left to right?
 Clyde, empty, Dutch, Gotti, Bugsy, Nelson
 Bugsy, Clyde, Nelson, Gotti, Dutch, empty
 Dutch, Bugsy, Gotti, Nelson, empty, Clyde
 Dutch, Clyde, Gotti, empty, Nelson, Bugsy
 Bugsy, Nelson, Gotti, Clyde, Dutch, empty
Solution
This problem is best solved by the method of elimination. To apply this method take a condition; test it against each answerchoice, eliminating any that violate it. Then take another condition; test it against the remaining answerchoices, eliminating any that violate it. Continue until only one answerchoice remains. Many students apply every condition to the first answerchoice, then every condition to the second answerchoice, and so on. This should be avoided since it’s inefficient; however, sometimes there is no other option. Because this question type is relatively easy, it often is the first or second question asked.
The first condition contradicts choice (A) since Clyde cannot be first. It does not contradict the other choices. So eliminate (A) only.
The second condition contradicts choice (B) since Gotti must be 3rd. It does not contradict the remaining choices. So eliminate (B) only.
The third condition contradicts choices (C) and (D) since in neither choice is Bugsy to the immediate left of Nelson.
It does not contradict the remaining choice. So eliminate (C) and (D) only.
Thus, by process of elimination, we have learned the answer is (E).
To answer this question, we had to test all the conditions; often, however, we will find the answer before testing the last condition.
The first condition contradicts choice (A) since Clyde cannot be first. It does not contradict the other choices. So eliminate (A) only.
To answer this question, we had to test all the conditions; often, however, we will find the answer before testing the last condition.
Question3
If space 6 is empty, which one of the following must be false?
 Clyde stands in space 4.
 Dutch stands in space 4.
 Clyde is to the left of Nelson.
 Clyde is to the right of Dutch.
 Nelson stands in space 2.
Solution
The structure of this question is awkward—the correct answer will always make a false statement! The question is more tractable when rephrased as “All of the following could be true EXCEPT.” Now, merely test each answerchoice against the initial conditions until you find the choice that violates one or more conditions.
Adding the supplementary condition, “space 6 is empty,” to the original diagrams gives
In Diagram I, the only space open for C is space 2:
___ C ___ G ___ X
Clearly, this diagram does not leave room for the condition BN. So we eliminate Diagram I.
Next, test each answerchoice against Diagram II, starting with (A). Place Clyde in space 4 as follows:
___ ___ G C ___ X
Now the condition BN forces B and N into spaces 1 and 2, respectively, which in turn forces D into space 5. So our uniquely determined diagram is
B N G C D X
This diagram does not violate any initial condition. Hence, Clyde could stand in space 4. So eliminate choice (A).
Next, turning to choice (B), place Dutch in space 4:
~C ___ G D ~C X
The condition BN forces B and N into spaces 1 and 2, respectively:
B N G D ___ X
But this diagram forces C into space 5, violating the condition C ≠ 1, 3, 5. Hence, D cannot stand in space 4, and the answer is (B).
The above method of analysis is what mathematicians and logicians call an “indirect proof”. To apply the method, assume an answerchoice is possible. Then check whether that leads to the desired result. If not, eliminate it. Then choose another answerchoice and repeat the process. Continue until either the choice with the desired result is found or until only one remains. This method of elimination is not as efficient as the previous one, because typically every condition must be tested against one answerchoice before considering the next answerchoice. Sometimes, however, this is the only method available.
Adding the supplementary condition, “space 6 is empty,” to the original diagrams gives
Diagram I ~C ___ ~C G ~C X
Diagram II ~C ___ G ___ ~C X
In Diagram I, the only space open for C is space 2:
___ C ___ G ___ X
Clearly, this diagram does not leave room for the condition BN. So we eliminate Diagram I.
Next, test each answerchoice against Diagram II, starting with (A). Place Clyde in space 4 as follows:
___ ___ G C ___ X
Now the condition BN forces B and N into spaces 1 and 2, respectively, which in turn forces D into space 5. So our uniquely determined diagram is
B N G C D X
This diagram does not violate any initial condition. Hence, Clyde could stand in space 4. So eliminate choice (A).
~C ___ G D ~C X
The condition BN forces B and N into spaces 1 and 2, respectively:
B N G D ___ X
But this diagram forces C into space 5, violating the condition C ≠ 1, 3, 5. Hence, D cannot stand in space 4, and the answer is (B).
The above method of analysis is what mathematicians and logicians call an “indirect proof”. To apply the method, assume an answerchoice is possible. Then check whether that leads to the desired result. If not, eliminate it. Then choose another answerchoice and repeat the process. Continue until either the choice with the desired result is found or until only one remains. This method of elimination is not as efficient as the previous one, because typically every condition must be tested against one answerchoice before considering the next answerchoice. Sometimes, however, this is the only method available.
Question4
Which one of the following spaces CANNOT be empty?
 1
 2
 3
 4
 5
Solution
Assume that space 1 is empty. Then in Diagram I, the condition BN can be placed in spaces 2 and 3, D can be placed in space 5, and C can be placed in space 6—all without violating any initial condition:
X B N G D C
Thus space 1 could be empty. This eliminates (A).
Next, assume that space 2 is empty. In Diagram I, this forces BN into spaces 5 and 6:
___ X ___ G B N
However, this diagram does not leave room for C (recall C ≠ 1, 3, 5). Diagram I is thus impossible when space 2 is empty. Turning to Diagram II, we see immediately that space 2 cannot be empty, for this would make G second, violating the condition G = 3rd. Hence, Diagram II is also impossible when space 2 is empty.
Thus space 2 cannot be empty, and the answer is (B).
X B N G D C
Thus space 1 could be empty. This eliminates (A).
___ X ___ G B N
However, this diagram does not leave room for C (recall C ≠ 1, 3, 5). Diagram I is thus impossible when space 2 is empty. Turning to Diagram II, we see immediately that space 2 cannot be empty, for this would make G second, violating the condition G = 3rd. Hence, Diagram II is also impossible when space 2 is empty.
Question5
If Clyde stands in space 6, Dutch must stand in space
 3 or 4
 5 or 6
 1 or 2
 2 or 3
 4 or 5
Solution
Adding the new condition, C = 6th, to the original diagrams yields
In both diagrams, BN must come before G, and D must come after G, to insure that G is 3rd. This forces D into space 5 in Diagram I and into either space 4 or 5 in Diagram II. In either position, D does not violate any initial condition.
Hence, the answer is (E).
Diagram I ___ ___ ___ G ___ C
Diagram II ___ ___ G ___ ___ C
In both diagrams, BN must come before G, and D must come after G, to insure that G is 3rd. This forces D into space 5 in Diagram I and into either space 4 or 5 in Diagram II. In either position, D does not violate any initial condition.
Points to Remember
 To apply the method of elimination, take a condition. Then test it against each answerchoice, eliminating any that violate it. Then take another condition; test it against the remaining answerchoices, eliminating any that violate it, and so on, until only one answerchoice remains.
 To apply the method of “indirect proof,” assume that a particular answerchoice is possible. Then check whether that leads to the desired result. If not, eliminate it. Then assume that another answerchoice is possible and repeat the process. Continue until the answerchoice with the desired result is found or until only one remains.