**Basic Ordering ** **Game**

The following is a free explanation of a basic Ordering Game. This game can be found as the first game in the LSAT Analytical Reasoning Section 1 from the June 2007 exam.

**STOP** Before you read this get your self a copy of the June 2007 LSAT and do the Analytical Reasoning section. (Free Copy of June 2007 LSAT Here) Then come here for the explanation of how to do the games. Book mark this page and come back when you have completed the June 2007 games section. Click here for some tips on **making the most of your practice LSATs**.

OKAY – so you have completed the Analytical Reasoning section of the June 2007 LSAT on your own. Now here is a full set of explanations for you to practice from.

A written explanation follows or you can check out the diagram set up and rules in this video:

At the top of the first page of Section one of the June 2007 LSAT you find the title Directions with an explanation of how to do this section. You should read and understand this now but DO NOT read it during the actual LSAT. The directions do not change and this is a waste of 30 seconds. It may not seem like much but in the end it could mean the difference between getting one more question right or one more wrong; and this could mean the difference between getting into your law school of choice or not. Read it now, read it tomorrow but DO NOT read it on the actual LSAT.

Let’s go on to the first game. Each game consists of an introductory paragraph, a set of rules and then a series of questions. As you read the introductory paragraph and set of rules you MUST summarize them all into a simple diagram and if necessary a few short hand rules on the side. We suggest a short hand for you to use here or you can develop your own, just make sure it is neat and easy to understand.

IMPORTANT – when you have completed summarizing the game and rules you must have included everything. You should not have to return to the rules or the introductory paragraph at any time when you are doing the questions. Your diagram and shorthand rules should give you all the information you need. If you find yourself returning to the game description you need to work on your diagram and shorthand rules notations.

Now let’s look at the introductory paragraph which describes the basic set up of the game. This is an Ordering game because you are placing entities in a single line in a particular order, or in other words you are sequencing entities in a line. The entities in this game are the digits. Remember that we must summarize all the information presented here. The key information here is that there are 5 (FIVE) digits in each code. Therefore, our diagram starts like this:

____ ____ ____ ____ ____

Five lines representing the five places to put our entities.

Next we read our rules and add them to the diagram or if they do not fit easily into the diagram then express them as a shorthand rule. Wherever possible try to fit them into the diagram.

Rule 1: Tells us that the digits are 0,1,2,3,4 and no others. There will almost always be a list of entities associated with any game. Write your list down above your diagram. It should now look like this:

0 1 2 3 4

____ ____ ____ ____ ____

REMEMBER – keep your diagram and shorthand neat. Messy diagrams can cost you lots of wrong answers. Those extra seconds to ensure your ones looks like a one and not a seven are important and well worth it.

Rule 2: Tells us that each digit is used exactly once in each code. This means that it is a perfect fit game. In a perfect fit game you have exactly the same number of different entities as you have places to fill and each is only used once. Here we have five places to fill for each code and five different entities with which to fill them. This means you can’t have two 3s or two 1s etc. in any given code. So we could not have 0 2 3 1 2 as a code. There is no need to include this rule in the diagram but if you like for now you can write down:

(Once each) next to the list of entities.

Rule 3: Tells us that the second digit is double the first digit. This can be added to our diagram as follows:

0 1 2 3 4

____ ____ ____ ____ ____

x2=

Meaning that the second digit is two times the first.

Rule 4: Tells us that the third digit must be less than the fifth digit. We include this as follows:

0 1 2 3 4

____ ____ ____ ____ ____

x2= < 5th > 3rd

The above diagram and list of entities includes everything we need to know about this game. It is all here. DO NOT go back and read the rules or introductory paragraph again. Note that you must be thorough in your diagram and rules because if you miss something it will be missing through out all your questions. Include it all the first time through.

Now that we have our complete diagram we can go on to the questions. BUT FIRST – the best LSAT writers will take a moment to expand upon the diagram if possible. This involves taking a quick look to see if there is anything else we can add. HOW DO YOU DO THIS? Everyone talks about it but what is going on in the mind of a good LSAT writer when they add to the diagram without any new rules. They think about the following:

- Combine Rules: Can any of my rules be combined to form a comprehensive rule? (none in this game)
- Limited Options: Are there a limited number of options as a result of the rules? In this game there is! How do we know this? Look at our diagram and see if a limited number of entities could satisfy each rule. The first rule in our diagram is that the second digit is exactly 2 times the first. This means that we can not have 0 or 3 or 4 as the first digit because we do not have a digit to put in the second spot that will be double the first digit. So, only 1 or 2 can go in the first spot, and this also means that only 2 or 4 can go in the second spot (being double the first)

This means we can add the following to our diagram:

0 1 2 3 4

1/2 2/4 ____ ____ ____

x2= < 5th > 3rd

You can use a slash mark to indicate that it is either one entity or another. So “1/2” means it is either 1 or 2 in the first spot. Note that if it is one in the first spot then two must be in the second spot. (In order to be double the first spot)

- Fill in Limited Options: A further step that works on many very difficult games is to take your limited options and fill them out as much as possible.

Note: What follows here is far more complex than is necessary for this game. If you are new to games we recommend you skip ahead and try this later when you are ready for some more advanced game techniques.

This technique is usually not necessary for simpler games like this one and can be a bit of a time waster but if you are struggling it can simplify the game for you and it definitely helps with more difficult games. It also is a good learning tool for practice.

To fill in the limited options just draw two new diagrams and fill them in as much as possible with the two options of 1 or 2 in the first spot. Here is how you can do this:

Option 1: 1 2 0/3 0/3/4 3/4

x2= < 5th > 3rd

Note that in Option 1 you can not have 4 in the 3rd spot as the digit in the 5th spot must be less than the digit in the 5th spot.

Option 2: 2 4 0/1 0/1/3 1/3

x2= < 5th > 3rd

**Back to the Game –** Now we have our complete diagram. Note that you should have this one master diagram that you do not touch again unless you are adding a new rule to the master diagram. The only reason you add a rule to the master diagram is when you find a rule that applies always. For example if you find out that 4 is always in the 5th spot (it is not). Other than that do not touch your master diagram, just redraw it for each question.

**Master Diagram:**

0 1 2 3 4

1/2 2/4 ____ ____ ____

x2= < 5th > 3rd

With our master diagram complete let’s move on to the questions. Each question is composed of a question stem which asks the actual question and the answer choices from A to E. If you have difficulty determining what the question stem is asking you to do see our **question stem guide**.

Question 1:

This is a **Must Be True** question. This question stem also gives us some extra information. This is in the part of the question that says “If the last digit of an acceptable product code is 1” Take this extra information and plug it into your diagram

You can follow along with the written explanation below or check out our video explanation:

1/2 2/4 ____ ____ __1__

x2= < 5th > 3rd

When we plug in 1 in the 5th spot we see that 1 can not be in the 1st spot. Remember that you can only use each entity once in this game. Therefore 2 must be in the 1st spot and from our rules we see that 4 must be in the 2nd spot as double the 1st spot. So we have:

2 4 ____ ____ __1__

x2= < 5th > 3rd

Don’t stop here, keep plugging in as many entities as possible without breaking the rules of your game. From our rules the 3rd digit must be less than the 5th digit. We only have two digits left: 0 and 3. So 0 must be in the 3rd spot and 3 in the 4th spot.

2 4 0 3 __1__

x2= < 5th > 3rd

We have completely filled in our diagram without even moving on to the answer choices. This will make the answer easy to find. We are looking for what Must Be True. Everything in our completed diagram for question 1 must be true, so just look for the answer choice that matches something in this diagram.

Answer (A) matches our diagram perfectly. 2 must be in the 1st spot.

All of answer choices (B) through (E) do not match our diagram, therefore they do not fit our Must Be True criteria. In fact in this question they cannot be true.

Question 2:

**IMPORTANT** – when you move on to a new question you must start back with your master diagram. Any additional rules imposed by previous questions is not included in subsequent questions. For example, in Question 1 we were told that the last digit of the product code was 1. This does not apply in the other questions. The last digit might be one but it does not have to be.

**TIME SAVING TIP** – you can still reuse filled in diagrams from previous questions. So if you have filled in a diagram of one possible correct and complete product code in a previous question you can refer to this for future questions. For example if a question asks if it can be true that 3 is in the 4th spot we can look back to our complete diagram from Question 1 and see that it is indeed possible to put 3 in the 4th spot. This is another good reason to keep your diagrams neat and also to not erase them.

This is a **Must Be True** question. However, unlike question 1 we are not given any additional information in the question stem so we must work only with our rules and master diagram as we have them. For Must Be True questions you should redraw a diagram for each answer choice and test each answer choice to see if it can be false. If it can be false then it is not the correct answer and we can cross it off. By this process of elimination we find the answer choice that must be true.

You can follow along with our written explanation below or check out our video explanation here:

(A)

Remember from our time saving tip above that we can refer back to previous diagrams for answering questions. We can see from Question 1 that the following is a complete and accurate sequence of entities:

2 4 0 3 __1__

x2= < 5th > 3rd

Note that digit 1 does not appear before digit 2. Remember that in a Must Be True question we want to try to prove that each answer choice can be false and then we can cross if off as a wrong answer and move on. As we see here it can be false that 1 is always before 2 and so we can cross off answer (A).

(B) Answer choice (B) can also be crossed off by looking at our diagram from Question

(C) For (C) we have to draw a new diagram. Let’s try to put 2 after 3 which if possible will allow us to cross this answer choice off and move on.

Redraw your diagram:

1/2 2/4 ____ ____ ____

x2= < 5th > 3rd

We know that 2 must either be in the 1st or second spot. Neither of these options allows us to put 2 after 3 because if we put 2 in the 2nd spot 1 must go in the 1st spot to satisfy our rule about the 2nd being double the 1st and if we put 2 in the first spot then there is no place before 2 to put 3. Therefore it Must Be True that 2 comes at some point before 3 and answer choice (C) is the correct answer.

Just to show you how the other answers need not be true or can be false:

(D) 1 2 0 3 4

(E) 1 2 0 3 4

Both of the above combinations meet all of our rules and show us that for (D) we need not have 3 before 0 and for (E) we need not have 4 before 3.

**Question 3**:

Here we have yet another **Must Be True** question. This one gives us the additional information that the third digit is not 0. Add this to each re-drawn diagram ONLY FOR THIS question.

1/2 2/4 ____ ____ ____

x2= < 5th > 3rd

not 0

Then plug in as much other information as possible. Unless you have taken the additional step of **Filling in your Limited Options** then there is no additional information or entities that you can plug in. If you did Fill in the Limited Options **See below ** for more details on how you can further complete this diagram.

**Limited Options: **

When we know that 0 cannot be in the 3rd position we can fill in our two limited options as follows: (for how we got our two limited options click here)

Option 1: 1 2 3 0 4

x2= < 5th > 3rd

Option 2: 2 4 1 0__ _3__

x2= < 5th > 3rd

**Back to the Game… **

Even if you did not go the extra step to **Fill in Limited Options** in the game set up you may also recognize that there are a very limited number of options if 0 cannot be in the 3rd spot. From our basic diagram 0 cannot be 1st or 2nd and now it cannot be 3rd. Also since the 3rd spot must be less than the 5th there is no way that 0 can be 5th as we have no digits less than 0. So 0 must be 4th.

Now let’s use the **Must Be True** method to find the correct answer choice.

(A) Can the second digit be something other than 2? Redraw the diagram: Remember that 0 cannot be 3rd for Question 3.

1/2 2/4 ____ ____ ____

x2= < 5th > 3rd

not 0

Now try to put 2 somewhere other than 2nd. The only other place 2 can go is 1st. Next see if we can fill in the diagram without breaking any rules, and also our new rule for this question that 0 cannot be 3rd.

2 4 ____ ____ ____

x2= < 5th > 3rd

not 0

First we cannot put 0 third so it must go 4th or 5th. But the 3rd digit must be less than the 5th so 0 cannot go 5th and must go 4th.

2 4 0 __

x2= < 5th > 3rd

not 0

The only two digits left are 1 and 3 and the only two spots left are 3rd and 5th. 3rd must be less than 5th so we have:

2 4 1 0 3__

x2= < 5th > 3rd

not 0

Here we have a complete diagram that satisfies all our rules and shows that 2 need not be second when 0 is not 3rd. Therefore answer choice (A) need not be true and so it is wrong. Cross if off and move on.

(B) Here is where you can save lots of time on the LSAT. Notice that 3 is not 3rd in our diagram for answer choice (A). No need to redraw a whole diagram and figure this out again. Cross off (B) and move on.

(C) We have already determined that this Must Be True. If 0 cannot be 3rd it must be 4th.

If you want you can even add this information to your master diagram. For all questions 0 must either be 3rd or 4th. It cannot go anywhere else. If you missed this extra rule don’t worry, it’s not necessary.

0 1 2 3 4

1/2 2/4 (0) (0) ____

x2= < 5th > 3rd

We use brackets (0) around an entity to express that these are the only options for where it can appear but that we do not know for sure which spot it appears in.

Just to show you that the answer choices (D) and (E) need not be true:

(D) 1 2 3 0 4

and

(E) 2 4 1 0 3

Show us that neither (D) nor (E) must be true and so they are wrong answers.

Question 4:

This question asks us which of the answer choices does not work. All will work EXCEPT one. To find the one that does not work, the correct answer, plug each answer choice into the diagram and test to see if it works. First let’s redraw our diagram. Remember we can include our new rule from Question 3. If you missed this don’t worry it’s not necessary.

0 1 2 3 4

1/2 2/4 (0) (0) ____

x2= < 5th > 3rd

Now let’s plug in the answer choices to see if they work:

(A) For answer choice (A) we put 0, 1 in the 3rd and 4th spots.

1/2 2/4 0 1 ____

x2= < 5th > 3rd

This means 1 cannot be 1st.

2 4 0 1 ____

x2= < 5th > 3rd

And then the only digit left for the 5th spot is 3.

2 4 0 1 __3__

x2= < 5th > 3rd

So this answer choice is possible and we can cross it off and move on.

(B) For answer choice (B) we put 0, 3 in the 3rd and 4th spots.

1/2 2/4 0 3 ____

x2= < 5th > 3rd

If we select 2 and 4 for the 1st and 2nd spots we have:

2 4 0 3 ____

x2= < 5th > 3rd

And then the only digit left for the 5th spot is 1. 1 is greater than 0 in the 3rd spot so we are okay by that rule.

2 4 0 3 __1__

x2= < 5th > 3rd

So this answer choice is possible and we can cross it off and move on.

(C) For answer choice (C) we put 1, 0 in the 3rd and 4th spots.

1/2 2/4 1 0 ____

x2= < 5th > 3rd

Since 1 is used in the 3rd spot we must have 2 and 4 in the 1st and 2nd spots so we have:

2 4 1 0 ____

x2= < 5th > 3rd

If we select 2 and 4 for the 1st and 2nd spots we have:

2 4 1 0 ____

x2= < 5th > 3rd

And then the only digit left for the 5th spot is 3.

2 4 1 0 __3__

x2= < 5th > 3rd

So this answer choice is possible and we can cross it off and move on.

(D) For answer choice (D) we put 3, 0 in the 3rd and 4th spots.

1/2 2/4 3 0 ____

x2= < 5th > 3rd

If we select 2 and 4 for the 1st and 2nd spots we have:

2 4 3 0 ____

x2= < 5th > 3rd

And then the only digit left for the 5th spot is 1. However, 1 cannot be in the 5th spot as the 5th spot must be greater than the 3rd spot and 3 is in the 3rd. So let’s go back and try our other option. Let’s put 1 and 2 in the 1st and 2nd spots:

1 2 3 0 ____

x2= < 5th > 3rd

The only digit left for the 5th spot is 4. Since 4 is greater than 3 in our 3rd spot this satisfies our rule.

1 2 3 0 __4__

x2= < 5th > 3rd

So this answer choice is possible and we can cross it off and move on.

(E) For answer choice (D) we put 3, 4 in the 3rd and 4th spots.

1/2 2/4 3 4 ____

x2= < 5th > 3rd

We can see right away that with 3 in the 3rd spot we need a greater number in the 5th spot. 4 is already taken and there is no other digit greater than 3 to fit in the 5th spot so this answer choice does not work, or in other words, when you put 3, 4 in the 3rd and 4th spots ,respectively, you cannot create an acceptable product code. Therefore, answer choice (E) is the correct answer.

Question 5:

Another Must Be True question. This one does not give us additional information in the question stem. Once again we must test each answer choice to see if it is possible for it to be false, if it is then we cross it off and move on until we find the one that Must Be True.

**TIME SAVING TIP** – remember that you can refer back to diagrams from past questions to save time.

(A) This one should be easy to find in past questions. Look at our diagram from Question 3 answer choice (A).

2 4 1 0 3__

x2= < 5th > 3rd

not 0

We see here that 0 and 1 do not need to have exactly one digit between them, here they have no digits between them. (A) is wrong, cross it off and move on.

(B) Look at our diagram from Question 1:

2 4 0 3 __1__

x2= < 5th > 3rd

Digits 1 and 2 need not be exactly one space apart. (B) is wrong, cross it off and move on.

(C) To test if this can be false let’s try to separate 1 and 3 by more than two digits. This would require them to be in the 1st and 5th spots. 3 cannot go first so let’s put 1 in the 1st spot.

1 2 ____ ____ __3__

x2= < 5th > 3rd

Then see if we can fill in the last two digits without breaking any rules. Only 0 and 4 are left. The 3rd spot must be filled by a digit less than 3 which is in the 5th spot. So we have:

1 2 __0__ __4__ __3__

x2= < 5th > 3rd

All our rules are satisfied here with 1 and 3 separated by more than 2 digits. Answer choice (C) can be false and so it is wrong. Cross (C) off and move on.

(D) From our diagram in question 4 (A) we have:

2 4 0 1 __3__

x2= < 5th > 3rd

Digits 2 and 3 can be separated by more than 2 digits. Answer choice (D) is wrong.

(E) Now we try to separate 2 and 4 by more than 2 digits. 2 is either in the 1st or second spot. If we put 2 in the first spot then 4 must be second and they are not separated by more than 2 digits.

2 4 ____

x2= < 5th > 3rd

If we put 2 in the second spot then there is no place where 4 can be more than 2 digits away from 2.

1 2 ____

x2= < 5th > 3rd

Therefore it Must Be True that 2 and 4 are separated by at most 2 digits.

>> The next step is to go over an advanced ordering game