re: It must be true that the lowest-numbered line on which You stated that, order could be: MGFSHKJ what about rule #3: H is assembled on line 1 or else line 7.

MGFSHKJ is not an option stated or referenced in the above (and only) question prompt that fits the description "It must be true that the lowest-numbered line on which". That is to say, as far as i can see, and as far as (ctrl + f) search of the page shows. Am I Wrong?

On the last practice question, why cant F be in the fifth spot? The rules are that M and G are a pair and that F goes before J. Can't K just go in the third spot and the order be MGKSFJH?

F can be in the 5th spot. As can J and K. They left it alone as all three can go there.

H is assigned on lines 1 or 7 could be written as H = 1/7

True ones with a strong math background would think to use fractions. But it is not always the case

Also for the question regarding K being in spot 2, it would have been easier to build scenarios around H rather than MG...the original rule forced H into spots 1 or 7, scenario#1 reads HKFSJMG and scenario #2 reads FKJSMGH...LEADS YOU TO ANSWER MUCH QUICKER AND EASIER THAN BUILDING AROUND MG SCENARIO

Actually, there is a scenario #3... HKFSMGJ... there are three total scenarios, and one could argue that writing out 3 scenarios might more time consuming than builing around the 2 MG scenarios

For the last question in the "How to approach ordering setups" article, I believe... Position 3 could be F or K Position 5 could be F, J or K Position 6 could be J or K If I am correct, why can't B be correct?

Hi there, Because if we're saying that FK are in positions 5 and 6 respectively, then J would have to be in 3rd - which would violate the F-J condition!

Main content

Course: LSAT > Unit 1

Lesson 3: Analytical Reasoning – Articles

How to approach ordering setups

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Ordering setups overview

When you see an ordering setup on Test Day (and you'll see at least one), it may be a great opportunity to be efficient and save some time—but only if you've practiced a specific approach that works for you. In this article, we’ll examine classic ordering setups (as opposed to mixed setups that include an ordering component), and share some tips to raise your game and boost your score.

How do we recognize ordering setups?

When the task involves the arrangement of elements—in other words, when order matters—then you’re working with an ordering setup.

A few examples of ordering tasks:

Determine the order of interviews scheduled from 1:00 to 6:00
Determine the seating arrangements of people around a circular table
Determine which parking spaces cars occupy, where the parking spaces are numbered 1 through 7
Determine what ranking (lowest to highest) six movies are given
Determine when five different dogs are adopted (Monday through Friday)

If order doesn’t matter in any aspect of the task, then you’re looking at a grouping setup, not an ordering setup. And if order matters for some parts of the task, but there are also elements to the task in which order doesn’t matter, then you’re looking at a mixed setup.

Grouping setups are covered in this article.

Mixed setups are covered in this article.

What should we pay attention to in the passage?

In the passage for an ordering setup, you’ll be given a certain number of elements and told that they'll be arranged in some way. Usually, the numbers will be very simple—six people and six spots, for example. Sometimes, however, the numbers won’t line up exactly that way; perhaps there are seven people interviewed over six days, where two people are interviewed on one day and one person is interviewed on each of the other days.

It's always a good idea to pay attention to how the number of elements relates to the number of spots!

What does the sketch for an ordering setup look like?

In general, it’s good to make a diagram to represent what’s happening in the setup. So, suppose we have a task in which persons ABCDEF need to be present on days Monday through Saturday. Your empty diagram (before establishing any rules or deductions) might look like this:

Note: It can be useful to make a list of all the “players” or “elements” in the setup. In the diagram above, and in many of the diagrams in Official LSAT Practice on Khan Academy, you’ll see these “player banks”—they can help you keep track of the elements you have used as you add them to the diagram.

In some cases, it makes sense to make a vertical sketch instead of a horizontal sketch. For example, if the six people worked in six different floors of a building, it would make sense to make the sketch resemble the floors of a building:

For consistency’s sake, we use the horizontal sketch as a default in our exercises, unless it makes more sense to build the sketch vertically (as in the building example above).

What kind of rules are typical in ordering setups?

Since ordering involves arrangement, some of the most common rules will involve how elements relate to each other. Suppose we have a task in which persons ABCDEF need to make presentations on days Monday through Saturday, one per day. We will likely see rules that resemble the following:

A presents on an earlier day than B.
A presents on the day immediately before B.
A cannot present immediately before B.
A presents sometime earlier than both B and C.
A presents on Monday.
A presents on either Monday or Thursday.
A and B’s presentations are separated by exactly two presentations.
If A presents on Thursday, then B presents earlier than F.

What are some good ways to represent these typical rules, and what can we deduce from each one?

It’s a good idea to get into a habit of notating rules consistently. That doesn’t mean that any one notation is the “correct” one, though! As we work through these rules, we’ll give you one or two ideas for each rule, and you can either use our idea or come up with your own. We’ll also let you know if there are any

that you can usually make for each rule, so that you can start to formulate deductions quickly and easily.

For the first two examples, we’ll show you the notation for a horizontal diagram on the left, and a vertical diagram on the right. For the remaining examples, we’ll show you how we’re notating the relationship in a horizontal ordering setup, and you can infer what the corresponding vertical-diagram notation would look like.

A presents on an earlier day than B.

In the vertical sketch on the right, we’re assuming that higher is earlier, and lower is later. This is a good example of when a horizontal sketch is likely easier to associate with “earlier/later”.
Both the ellipses and the hyphen are acceptable, but when it looks like you’ll be continuing on to create a “chain” of elements that branch off from each other, the hyphens can be quicker and easier.
Don’t make the mistake of inferring that A must be immediately before B!

Deductions:

In our setup, we could immediately deduce that A cannot present on Saturday, and B cannot present on Monday. That’s because if we tried to have A present on Saturday, there would be no room for B to be later!
A similar conflict occurs when we try to have B present on Monday.
Since you’re tested as much on what’s true as on what’s false, it can be helpful to make this deduction as soon as you note the rule; we can note it directly into our basic diagram:

A presents on the day immediately before B.

We often call this a “pair.” If there are more than two consecutive elements, we’ll call it a “group.”
Since no elements can ever come between A and B and we therefore know exactly what their relationship is to each other, this pair is more restricted than an A...B relationship would be. In other words, there are fewer possible placements for where an AB pair can go than for an A...B relationship. The AB pair requires two consecutive spots, whereas the A...B relationship does not.

Deductions:

As with the A...B relationship, we should immediately deduce that A cannot present on Saturday, and B cannot present on Monday. We can note it directly into our basic diagram in the same way.

A cannot present immediately before B.

Deductions:

Since this rule tells us about something that can’t happen, we can’t make any useful deductions from this rule alone. What’s likely is that either:
1) this rule will be combinable with another rule in order to make deductions (for example, a rule that states, “B presents on Tuesday” would allow us to deduce that A can’t be on Monday), or
2) in the course of the questions, either A or B will be established by a new condition, and we’ll be able to deduce that the other of the illegal pair can’t be adjacent.

A is sometime earlier than both B and C.

Deductions:

Since the rule tells us that A is sometime before at least two people, we can deduce that A can’t be Friday or Saturday.
Since A is earlier than B, we can deduce that B is not on Monday.
Likewise, since A is earlier than C, we can deduce that C is not on Monday.
Again, we don’t know the relationship between B and C!

A presents on Monday.

Deductions:

While the rule at face value may seem obvious, it’s worth noting that because we’ve established an element, one of the spots is essentially “not in play” anymore. That means that there are fewer places that the other entities can occupy.

A presents on either Monday or Thursday.

Deductions:

Since A is either Monday or Thursday, that means A is not Tuesday, Wednesday, Friday, or Saturday. Some students like to “double up” on the notations and not only mark A with the arrows to Monday or Thursday but also mark the not A under the other days.
In the questions, pay attention to A! If you’re given a new condition, for example, “F is on the day before A”, then you’ll be able to deduce that A must be on Thursday in that case.

A's presentation and B’s presentation are separated by exactly two presentations.

Deductions:

Since we don’t know the order of A and B, we can’t make deductions about where A and B can’t go.
If the rule had stated that A and B are separated by exactly two presentations and that A is earlier than B, then we could have deduced that A isn’t Thursday, Friday, or Saturday (so Wednesday at the latest) and that B isn’t Monday, Tuesday, or Wednesday (and so Thursday at the earliest).
Note that this relationship is very restricted in where it can go. Either A and B are Monday/Thursday, or A and B are Tuesday/Friday, or A and B are Wednesday/Saturday.

If A presents on Thursday, then B presents earlier than F.

Deductions:

It’s important to remember that conditional rules also have logically equivalent rules built into them: if B doesn’t present earlier than F, then A doesn’t present on Thursday.
Since there can’t be two presenters on the same day, saying that B doesn’t present earlier than F is exactly the same as saying that F presents earlier than B!
It’s crucial to be aware that we can’t deduce anything if A doesn’t present on Thursday.
So, underneath the original rule, you should note the equivalent rule:

Practice Example

Now, let’s take everything we’ve learned and apply it to an actual LSAT setup!

Exactly seven toy-truck models—F, G, H, J, K, M, and S—are assembled on seven assembly lines, exactly one model to a line. The seven lines are arranged side by side and numbered consecutively 1 through 7. Assignment of models to lines must meet the following conditions:

F is assembled on a lower-numbered line than J.

M is assembled on the line numbered one lower than the line on which G is assembled.

H is assembled on line 1 or else line 7.

S is assembled on line 4.

What does the setup tell us?

The passage is a typical ordering setup. There are seven toy-truck models and seven assembly lines, so the numbers line up as we would expect them to. Go ahead and make a basic sketch based on this information before we look at the rules!

You may have created something like this for your sketch:

Now, go ahead and note the rules, making sure to take the time to understand them! What do your notes look like?

What do the rules tell us?

Taking each rule one at a time, can you note each rule either directly in the diagram (if it makes sense to) or off to the side?

If you followed our notation suggestions, your notes could look like this. If you used your own and they make sense to you—great!

Rule 1: F is assembled on a lower-numbered line than J. This means that F will always be somewhere to the left of J, but we can’t be sure how far to the left.
Rule 2: M is assembled on the line numbered one lower than the line on which G is assembled. This means that M will always be immediately to the left of G.
Rule 3: H is assembled on line 1 or else line 7. This means that in any acceptable scenario, H will be in spot 1 or spot 7, but never anywhere else. This makes H rather restricted in comparison to the other models.
Rule 4: S is assembled on line 4. These are always the easiest rules, since they establish an element in a specific spot—we’ll never need to worry about where S goes.

Now it’s time to make deductions!

What can we deduce?

Can you use what we covered in this lesson in order to deduce what must be true beyond what’s presented to us at face value?

What can we deduce?

From Rule 1, we can deduce that F can’t be 7th and J can’t be 1st. These impossibility deductions are crucial to getting quick points on the test!
Likewise, from Rule 2 we can deduce that M can’t be 7th and G can’t be 1st. Notice that we aren’t concerned with where models can go—there are too many possibilities! Focus on what must be true and what must be false.
If Rule 4 didn’t exist, we would actually be done with the important deductions. However, we’re told that S must be 4th. So S is firmly established in the 4th spot, and no other model can ever be placed there.

Do you see another rule that you can combine with Rule 4 in order to determine what can’t happen?

If we combine Rule 4 with Rule 2, then we know that M can’t be 3rd! If we tried to place M 3rd, that would force G into the 4th spot. That conflicts with S in the 4th spot!
Similarly, G can’t be 5th. If we tried to place G 5th, then Rule 2 would force M to be 4th, and that’s a conflict with S as well.

Here’s an example of an initial diagram for this setup, based on what we uncovered:

We’re in a really great spot to move on to the questions! With this diagram, we won’t have to use so much (if any) of the “trial and error” approach that consumes a ton of time on Test Day.

We strongly recommend against using this initial diagram to “test” choices or sketch possibilities. It’s generally better for accuracy purposes (even if it takes a tiny bit more time!) to do a quick re-sketch when a question provides you with a new condition to consider. What this initial diagram represents is everything that’s always true, in every acceptable scenario. What “new condition” questions present are temporary conditions, so you likely don’t want to mix temporary conditions with permanent conditions. That said, you don’t need to resketch the entire initial diagram for each “new condition” question; a bare-bones version while you consult the original version can work great.

Which one of the following is an acceptable assignment of toy-truck models to lines, in order from line 1 through line 7?

(A) F, J, K, S, H, M, G
(B) F, K, J, S, M, G, H
(C) F, M, K, S, G, J, H
(D) H, K, S, M, G, F, J
(E) H, M, G, S, J, F, K

One approach that tends to be very efficient might seem a little counterintuitive at first. Instead of looking at each choice and evaluating it against all of the rules, it makes sense to look at each rule one at a time and eliminate the choices that violate that rule. Think about it this way: if the answer is (E), then looking at each choice first means going through all of the rules five times. But if you start with the rules and eliminate choices after looking at each rule, then you only have to go through the rules once!

What choice(s) does each rule eliminate when you evaluate each rule one at a time? Let’s take a look at them together!

Rule 1: F...J. (E) breaks this rule, so we can eliminate (E) and never consider that choice again.
Rule 2: MG is a pair. (C) violates this rule.
Rule 3: H is first or last. (A) doesn’t conform to this rule.
Rule 4: S is 4th. We can cross off (D).

The only choice remaining is (B), so that’s our answer!

It must be true that the lowest-numbered line on which

(A) F can be assembled is line 2
(B) G can be assembled is line 3
(C) J can be assembled is line 2
(D) K can be assembled is line 3
(E) M can be assembled is line 2

This question asks us to identify the choice that must be true, regarding the lowest-numbered line on which a model could be assembled. The wrong choices will be those that could be false; in other words, the model listed could be lower than the spot that the choice assigns it. Always take time to understand your task fully before diving in.

Can you consult your initial diagram and determine which statement must be true in terms of a model’s lowest possible line?

This is where our initial diagram will save us a lot of time! To test the choices effectively against your initial diagram in this question—usually without having to sketch anything out—ask yourself whether the statement in the choice has to be true. If you could make it not be true and still have an acceptable scenario that doesn't break any rules, then that choice qualifies as a could be false—not a must be true!

(A): Does it have to be true that F can be no lower than line 2? In other words, could F be 1st?

F could be lower than line 2! We have no rules stopping F from being 1st, and for completion’s sake—though you probably wouldn’t do this on Test Day—we’ll give you an example. F could be 1st, for example if the order were: FMGSJKH. Although there are several other possible scenarios in which F can be 1st, finding just one is sufficient to determine that this choice’s statement “could be false.”

(B): Does it have to be true that G can be no lower than line 3? In other words, let’s test to see if G could be 1st or 2nd.

Our diagram shows us that G can’t be 1st, so there’s no need to test that possibility.
Could G be 2nd? Sure, if M were 1st. If you want to verify this, since we’re not being timed right now, the order could be: MGFSKJH, to name just one possibility.

(C): Does it have to be true that J can be no lower than line 2? In other words, let’s test to see if J could be 1st.

Our diagram shows us that J can’t be 1st—we deduced this in our initial setup! Therefore, there’s no need to try anything out with our pencil. It must be true that J will never be assembled on a lower line than line 2!

On Test Day, you would have stopped and selected (C) as the answer. However, for completion’s sake, let’s look at (D) and (E) in case you have questions.

(D): Does it have to be true that K can be no lower than line 3? In other words, can K be 1st or 2nd?

K can be 1st, for example if the order were: KMGSFJH.

(E): Does it have to be true that M can be no lower than line 2? In other words, can M be 1st?

M could be 1st (and thus lower than line 2), for example if the order were: MGFSKJH.

If K is assembled on line 5, which one of the following is a pair of models that could be assembled, not necessarily in the order given, on lines whose numbers are consecutive to each other?

(A) G, H
(B) G, J
(C) H, J
(D) J, M
(E) M, S

For questions that give you a new, temporary condition to consider, it usually makes sense to re-draw a bare-bones sketch of your initial diagram and incorporate the new condition:

Now, can you use your initial diagram to make further deductions? In other words, what is the impact of K being placed into spot 5?

First, notice that we don’t have any rules regarding K—so the first obvious place to look for deductions doesn’t pan out.
Don’t despair, though! There’s always enough information to find the answer.
Remember when we said that H is a restricted element, because it can only be placed into two different spots? Let’s capitalize on that now, and build two scenarios around H!

Scenario 1: H is 1st.

Aha—notice that when we place H 1st, we’re left with two pairs of consecutive spots.
Rule 2 establishes the pair MG! So, either MG is 2-3 and FJ (the remaining models) is 6-7, or vice versa.
We can note those two pairs off to the side, since we don’t know which pair goes into which adjacent spots.

Scenario 2: H is 7th.

When H is 7th, the MG pair can only fit into spots 1-2 and spots 2-3.
That means that in either case, M or G must be in spot 2.
Now we can look at our list of models to see who’s unaccounted for. There’s a free spot sometime before S and a free spot sometime after S, and it so happens that we haven’t accounted for F and J yet.
F...J is what Rule 1 tells us! So,
F is either 1st or 3rd, and J must be 6th.
That means that either F or M is 1st, and either F or G is 3rd:

We’re ready to evaluate the choices!

Which choice contains models that could be consecutive to each other in some order?

(A): Can GH be consecutive in some order?

GH can’t be a pair in either scenario. In Scenario 1, H is 1st while G is either 3rd or 7th; in Scenario 2, G is either 2nd or 3rd, while H is 7th.

(B): Can GJ be consecutive in some order?

GJ can’t be a pair in either scenario. In Scenario 1, J and G are 3rd and 7th in some order, and in Scenario 2, G is either 2nd or 3rd, while J is 6th.

(C): Can HJ be consecutive in some order?

HJ could be a pair, because they’re a pair in Scenario 2. In that scenario, we deduced that H is 7th and J is 6th!

On Test Day, we’d stop here, but since we’re not being timed, let’s look at the remaining choices:

(D): Can JM be consecutive in some order?

JM can’t be a pair in either scenario. In Scenario 1, FJ must be one of the two pairs, and in Scenario 2, M is 1st or 2nd while J is 6th.

(E): Can MS be consecutive in some order?

MS can’t be a pair in this question. In Scenario 1, S is 4th while M is either 2nd or 6th, and in Scenario 2, S is 4th while M is either 1st or 2nd.

The answer is (C)!

There can be at most how many lines between the line on which F is assembled and the line on which J is assembled?

(A) one
(B) two
(C) three
(D) four
(E) five

Remember, we want to start with the highest number here, since we're looking for the maximum:

Check (E): can five spaces separate F and J?

For five spaces to separate F and J, one of them would have to be 1st and the other 7th. This doesn’t work, because H needs a place to go, and Rule 3 tells us that H must be 1st or 7th.

Check (D): can four spaces separate F and J?

It may help to make a quick sketch here. Since F is earlier than J (Rule 1), let’s make F 1st and J 6th.
That would make H 7th.
The only spots available for the MG pair are 2-3.
Only K is then available for the 5th spot.
This works! The order is: FMGSKJH.

On Test Day, you could select (D) as the answer and move on, if you're confident in your work.

If K is assembled on line 2, which one of the following must be true?

(A) F is assembled on a lower-numbered line than S.
(B) H is assembled on a lower-numbered line than G.
(C) J is assembled on a lower-numbered line than H.
(D) M is assembled on a lower-numbered line than J.
(E) S is assembled on a lower-numbered line than J.

Now, can you use your initial diagram to make further deductions? In other words, what is the impact of K being placed into spot 2?

First, notice that we don’t have any rules regarding K—so the first obvious place to look for deductions doesn’t pan out.
The next thing we could look or is whether H is forced into either 1st or 7th—but it’s not.
Next, our MG pair could come in handy. Given that MG could only be 5-6 or 6-7 in this question, we can again build scenarios around MG!

Scenario 1: MG is 5-6.

In this case, we have established K, S, M, G so we still need to account for H, F, J.
Since we deduced in our initial setup that J can’t be 1st, then either F or H must be 1st.
Since we similarly deduced in our initial setup that F can’t be 7th, then either H or J must be 7th.
Since H can’t be 3rd, then either F or J must be 3rd:

Scenario 2: MG is 6-7.

In this case, H is forced into the 1st spot.
That leaves only F and J to be accounted for, so F must be 3rd, and J must be 5th (since Rule 1 tells us that F...J).
This scenario is completely determined:

To find the answer, we’ll use these scenarios and identify the statement that must be true, because it’s true in both scenarios!

(A): Is F lower than S in both scenarios?

Yes! In Scenario 1, F is either 1st or 3rd, both of which are lower than S in 4th. And in Scenario 2, F is 3rd and S is 4th.

On Test Day, we’d be glad that the answer is (A) and move on to the next question. Here are the rationales for the remaining choices:

(B): This is true in Scenario 2, but it could be false in Scenario 1. There, H could be 7th, which would make it a higher-numbered line than G in 6th: FKJSMGH.
(C): This could be false, because it is false in Scenario 2. There, H is 1st and J is 5th, so J is assembled on a higher-numbered line than H.
(D): This could be false, because it is false in Scenario 2. There, J is 5th and M is 6th, so M is assembled on a higher-numbered line than J.
(E): This is true in Scenario 2, but it could be false in Scenario 1. There, J could be 3rd, which would make S assembled on a higher-numbered line than J: FKJSMGH.

(A) is the answer.

If G is assembled on the line numbered one less than the line on which F is assembled, then which one of the following must be true?

(A) F is assembled on line 3.
(B) G is assembled on line 5.
(C) H is assembled on line 1.
(D) K is assembled on line 5.
(E) M is assembled on line 6.

For questions that give you a new condition, it usually makes sense to re-draw a bare-bones sketch of your initial diagram and incorporate the new condition:

Now, can you use your initial diagram to make further deductions? In other words, what is the impact of G being placed into one spot lower than F?

First, notice G is also mentioned in the rule that gives us MG as a pair.
That means that MGF is a group!
How many places can the MGF group go? We have spots 1-3 open and spots 5-7 open to accommodate the group.
However, we deduced in our initial setup that F can’t be 7th. So the MGF group must be 1-2-3, respectively!
This means that H must be 7th, since it can’t be 1st.
J and K are 5th and 6th, but we don’t know in which order:

We should be able to quickly identify the answer now. Which one of the choices must be true?

To find the answer, we’ll use our new diagram for this question and identify the statement that must be true.

(A): Does F have to be assembled on line 3?

Yes! We deduced in this question’s setup that this statement must be true. If G is one spot lower than F, then the order is either MGFSJKH or MGFSKJH. In both cases, F is 3rd.

On Test Day, we’d move on efficiently to another question, without ever having had to use trial and error! In case you're curious about the wrong choices, which are statements that could be false:

(B): This not only could be false, it must be false. We deduced that in this question, G must be 2nd, not 5th.
(C): This not only could be false, it must be false. We deduced that in this question, H must be 7th, not 1st.
(D): This could be true, but it could also be false. We deduced that in this question, K could be 5th or 6th.
(E): This must be false. We deduced that in this question, M must be 1st, not 6th.

If M is assembled on line 1, which one of the following could be true?

(A) F is assembled on a line numbered one lower than the line on which H is assembled.
(B) F is assembled on a line numbered one lower than the line on which K is assembled.
(C) G is assembled on a line numbered one lower than the line on which J is assembled.
(D) G is assembled on a line numbered one lower than the line on which K is assembled.
(E) K is assembled on a line numbered one lower than the line on which G is assembled.

For questions that give you a new condition, it usually makes sense to re-draw a bare-bones sketch of your initial diagram and incorporate the new condition, which is that M is 1st:

Now, can you use your initial diagram to make further deductions? In other words, what is the impact of M being 1st?

First, notice M is also mentioned in the rule that gives us MG as a pair.
That means that G must be 2nd.
Therefore, H must be 7th, since it can’t be 1st.
Since F...J, we know that F can’t be 6th, so the 6th line is either J or K.
Likewise, J can’t be 3rd, so the 3rd line is either F or K.

We should be able to quickly identify the answer now. Which one of the choices could be true in our new diagram?

We can skim the choices to quickly find the one that's possible based on our initial diagram!

(D): Can G be assembled on a line numbered one lower than the line on which K is assembled?

Yes! We deduced that G is 2nd in this question, and the 3rd spot is either F or K. So G could be on the line immediately lower than K if the order is, for example: MGKSFJH.

On Test Day, we’d move on efficiently to another question, without ever having had to sketch anything out!

In this question, the wrong choices are statements that must be false (or in other words, are impossible).

(A): This must be false. We deduced that H is 7th, and the 6th spot can only be J or K. So F can’t be one lower than H.
(B): This must be false. F could be 5th, but only if J were 6th (because of the F...J rule).
(C): This must be false. We deduced that G is 2nd in this question, and the 3rd spot can only be F or K. So, GJ can’t be a pair in this scenario.
(E): This must be false. According to the setup rules, M (not K) must be one line lower than G.

(D) is the answer.

Takeaways

You recognize ordering setups by the fact that arrangement matters.
Read rules carefully! "Before" and "after" are easy to transpose if you're rushing.
Don't skimp on deductions! Every time you have an A...B rule or an AB rule, for example, you can make quick deductions as to where A and B can't go. Take the time to do so, because points come from deductions!
Practice hard! Since you will see an ordering setup on Test Day, you can practice what you've learned in this article until it becomes second nature.
Think twice before using trial-and-error. Trial-and-error sketching is not a time-efficient way to answer Analytical Reasoning questions. Make sure to draw a new sketch for questions that give you a new condition, and for those that don't, think hard about what the question is asking you to do.

Next steps

Watch a demonstration of how to approach an ordering setup

Then, once you've finished your diagnostics and created a practice schedule, head over to the practice area and try some ordering setups on your own!

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Kk W
Posted 5 years ago. Direct link to Kk W's post “For the examples given ho...”
For the examples given how would you rate their difficulty in reference to the LSAT , Basic,medium or Advanced
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(24 votes)
Answer
Carlos Garcia
Posted 6 years ago. Direct link to Carlos Garcia's post “re: It must be true that ...”
re: It must be true that the lowest-numbered line on which
You stated that, order could be: MGFSHKJ
what about rule #3: H is assembled on line 1 or else line 7.
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(13 votes)
Answer
- vincent cedre
  Posted 5 years ago. Direct link to vincent cedre's post “MGFSHKJ is not an option ...”
  MGFSHKJ is not an option stated or referenced in the above (and only) question prompt that fits the description "It must be true that the lowest-numbered line on which".
  That is to say, as far as i can see, and as far as (ctrl + f) search of the page shows. Am I Wrong?
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  (2 votes)
salazar.adriana1998
Posted 6 years ago. Direct link to salazar.adriana1998's post “On the last practice ques...”
On the last practice question, why cant F be in the fifth spot? The rules are that M and G are a pair and that F goes before J. Can't K just go in the third spot and the order be MGKSFJH?
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(4 votes)
Answer
- rohan9gupta
  Posted 6 years ago. Direct link to rohan9gupta's post “F can be in the 5th spot....”
  F can be in the 5th spot. As can J and K. They left it alone as all three can go there.
  Comment on rohan9gupta's post “F can be in the 5th spot....”
  (5 votes)
Leaferly
Posted a year ago. Direct link to Leaferly's post “For the fifth practice qu...”
For the fifth practice question, how do you decide to use the restricted element MG to build scenarios around, rather than using the restricted element H? My first thought was to use H, which led me to taking more time as it produced 3 scenarios instead of 2. Thanks!
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(4 votes)
Answer
andreastocks
Posted 4 years ago. Direct link to andreastocks's post “H is assigned on lines 1 ...”
H is assigned on lines 1 or 7 could be written as H = 1/7
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(2 votes)
Answer
- bennyb21
  Posted 4 years ago. Direct link to bennyb21's post “True ones with a strong m...”
  True ones with a strong math background would think to use fractions. But it is not always the case
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  (3 votes)
andreastocks
Posted 4 years ago. Direct link to andreastocks's post “Also for the question reg...”
Also for the question regarding K being in spot 2, it would have been easier to build scenarios around H rather than MG...the original rule forced H into spots 1 or 7, scenario#1 reads HKFSJMG and scenario #2 reads FKJSMGH...LEADS YOU TO ANSWER MUCH QUICKER AND EASIER THAN BUILDING AROUND MG SCENARIO
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(2 votes)
Answer
- Dan R
  Posted 4 years ago. Direct link to Dan R's post “Actually, there is a scen...”
  Actually, there is a scenario #3... HKFSMGJ... there are three total scenarios, and one could argue that writing out 3 scenarios might more time consuming than builing around the 2 MG scenarios
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  (3 votes)
carolynthomas705
Posted 2 years ago. Direct link to carolynthomas705's post “Why couldn't F go in the ...”
Why couldn't F go in the 6th spot?
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(2 votes)
Answer
marykathleen.lord
Posted 4 years ago. Direct link to marykathleen.lord's post “For the last question in ...”
For the last question in the "How to approach ordering setups" article, I believe...

Position 3 could be F or K
Position 5 could be F, J or K
Position 6 could be J or K

If I am correct, why can't B be correct?
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(2 votes)
Answer
- Zoe Wulff
  Posted 4 years ago. Direct link to Zoe Wulff's post “Hi there, Because if we'...”
  Hi there,
  
  Because if we're saying that FK are in positions 5 and 6 respectively, then J would have to be in 3rd - which would violate the F-J condition!
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  (2 votes)
Carah Conner
Posted 4 years ago. Direct link to Carah Conner's post “Im so confused, The initi...”
Im so confused, The initial rule was F being lower than J meaning that it goes after J. If that is correct then I may have gotten lost in translation to understand how E would be the wrong and answer.
Button navigates to signup pageComment on Carah Conner's post “Im so confused, The initi...”
(2 votes)
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andreastocks
Posted 4 years ago. Direct link to andreastocks's post “A presents on Monday coul...”
A presents on Monday could be written as A = M when writing rules instead of drawing barebones diagram and drawing A in Monday slot...takes up time
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(2 votes)
Answer

LSAT

Course: LSAT > Unit 1

Ordering setups overview

How do we recognize ordering setups?

What should we pay attention to in the passage?

What does the sketch for an ordering setup look like?

What kind of rules are typical in ordering setups?

What are some good ways to represent these typical rules, and what can we deduce from each one?

Practice Example

Beginning of reading passage.

Takeaways

Next steps

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