Pg. 381-2 Pure Sequencing (Mutually Exclusive Outcomes)

[bragg15]

Hi,

Thanks in advance for your help. I am working through the Pure Sequencing section, and on page 381-2 the book is discussing mutually exclusive outcomes.

The second example is giving me a tough time.

I understand how we arrived at the first two solutions of A > B > C and C > B > A, but I am not comprehending how the authors arrived at the additional two possible sequences of B > A > C and B > C > A.

Can someone help?

Thanks,
Marcus (Feb 2014 LSAT - Let's go!)

[Dave Killoran]

Hey Marcus,

Thanks for the question (and I like your signature slogan!).

The deductions you ask about are tricky ones, so let's take a moment to explore what's happening there. First, I'll bypass the A > B > C and C > B > A chains since you stated that they made sense. What's important about those is that they are each triggered by sufficient conditions, the first being A > B, and the second being C > B. When either of those two conditions are present, then one of the chains will be created. But what happens when it's B > A instead of A > B, or when it B > C instead of C > B? Let's take the case of B > A and see what occurs.

When B > A, the rule that A > B B > C is NOT enacted because the sufficient condition is not met. So, we're not violating any rule by proceeding and we can take a look at where C could appear. If B > A, what are our options with C? It could go in three places:

• 1. Earlier than B ( C > B > A )
  
  This option is just fine since it meets the conditions of the contrapositive (C > B). So, we're good here.
  
  
  2. Between B and A ( B > C > A )
  
  This is the first of the two chains that concerned you. First, are there any violations of the A > B B > C rule or its contrapositive? No, neither A >B or C >B are present, so the rule is not enacted and we don't have to worry about the necessary condition occurring. It's kind of like saying, "Ok, we have this chain B > C > A — is that possible under A > B B > C ? Yes, it is, because the rule begins with A > B, and we don't have that in the chain, so there is no conflict."
  
  
  3. Later than A ( B > A > C )
  
  This is the second chain that concerned you, and the logic behind how this works is identical to the logic in #2 above.

Note that you could approach these chains from the starting point of B > C (instead of B > A as we did above) and by placing A we'd find the exact same outcomes (although point #1 would yield the chain of A > B > C). In either event, the chains B > C > A and B > A > C don't violate the given rule, so they are ok. It's just really hard to see that they exist since the rule is so visually suggestive of A >B > C and C > B > A. It's a tough point, and one I'll clarify in the next edition of the book, so thanks for asking about it!

Please let me know if that helps. Thanks!

[bragg15]

Hey Marcus,

Thanks for the question (and I like your signature slogan!).

The deductions you ask about are tricky ones, so let's take a moment to explore what's happening there. First, I'll bypass the A > B > C and C > B > A chains since you stated that they made sense. What's important about those is that they are each triggered by sufficient conditions, the first being A > B, and the second being C > B. When either of those two conditions are present, then one of the chains will be created. But what happens when it's B > A instead of A > B, or when it B > C instead of C > B? Let's take the case of B > A and see what occurs.

When B > A, the rule that A > B B > C is NOT enacted because the sufficient condition is not met. So, we're not violating any rule by proceeding and we can take a look at where C could appear. If B > A, what are our options with C? It could go in three places:

• 1. Earlier than B ( C > B > A )
  
  This option is just fine since it meets the conditions of the contrapositive (C > B). So, we're good here.
  
  
  2. Between B and A ( B > C > A )
  
  This is the first of the two chains that concerned you. First, are there any violations of the A > B B > C rule or its contrapositive? No, neither A >B or C >B are present, so the rule is not enacted and we don't have to worry about the necessary condition occurring. It's kind of like saying, "Ok, we have this chain B > C > A — is that possible under A > B B > C ? Yes, it is, because the rule begins with A > B, and we don't have that in the chain, so there is no conflict."
  
  
  3. Later than A ( B > A > C )
  
  This is the second chain that concerned you, and the logic behind how this works is identical to the logic in #2 above.

Note that you could approach these chains from the starting point of B > C (instead of B > A as we did above) and by placing A we'd find the exact same outcomes (although point #1 would yield the chain of A > B > C). In either event, the chains B > C > A and B > A > C don't violate the given rule, so they are ok. It's just really hard to see that they exist since the rule is so visually suggestive of A >B > C and C > B > A. It's a tough point, and one I'll clarify in the next edition of the book, so thanks for asking about it!

Please let me know if that helps. Thanks!

Dave,

Thanks so much for responding! This makes perfect sense. As I understand it, as long as you don't violate the premise of the original conditional or its contrapositive, the variables should logically be placed elsewhere in the sequence to form the other possibilities.

When you write the next edition, explain it just the way you did here and they'll be sure to get it.

Thanks again,
Marcus

[lucia]

Hello,

I am going through Chapter 7 and really stuck on the last concept in the book, introduced on pgs 381-82.

This is illustrated through the statement "If A's presentation is earlier than B's presentation, then B's presentation is earlier than C's presentation."

I understand how to diagram this and I understand the contrapositive--the book explains that clearly:
A>B>C
C>B>A

What I can't seem to wrap my head around is the bottom of page 382, top of 383: "But, while both sequences are possible two other possible sequences exist:"

B>A>C
B>C>A

So... this makes sense. But what's the rule of thumb here? What I am guessing is that these two statements are a result of the necessary being true in the original and contrapositive statement and then the sufficient being false as a result from that. I thought about this like a normal conditional statement: given A -->B, the possibilities are: A and B, not B and not A, or B with or without A. BUT then in the diagramming drill that follows on the next few pages, these extra possibilities aren't considered in the answer choices in #3, #5, or #6. So I'm totally confused.

In sum: what's a good way to remember when to look for these additional cases? And, when you know to look for them--what's an easy way to know how to put them together correctly?

[Dave Killoran]

Hi Lucia,

Thanks for the question! The case you reference here is an interesting one, and as you've rightly noted, it revolves around the "IF A > B" segment of the rule. If A > B, then we know there are rules that have to be followed. If it's not the case that A > B, then we get the other possible options. I think that based on what you've said, that all makes sense to you.

The rule of thumb, then, is that we simply have to make sure to understand all options that include A > B (for when the sufficient condition is met, and of which there is just one: A > B > C) and all that include B > A (for when the sufficient is not met, and of which there are three: C > B > A, B > A > C, and B > C > A). Those are the same possibilities you reference when talking about A and B in the A B relationship (they just look different due to the greater complexity of the conditions here).

In the drills that follow, note that we are just diagramming the rules, not showing all the resultant possibilities. That's why #1, for example, which is identical to the relationship we are talking about here, just shows the diagram of the original statement and then the contrapositive. It doesn't draw out the other possibilities (because you often won't have time to draw out every variation that is possible under a rule; you simply have to know what the options are by looking at the rule and understanding the way it works). those possibilities still exist, and just having the discussion on pages 379-383 helps you know about them.

Please let me know if that helps out. Thanks!

[hkashi73]

Hi,

I'm struggling to understand why there would be 4 possibilities in the problem presented in page 422 chapter 7. I understand everything up until the 3rd and 4th possible sequences are introduced. How do we derive those two possibilities from the given example? In other words how do we infer the relationship between A and C and why does B stay ahead of A and C.

If someone could clarify this, I would appreciate it.

[Dave Killoran]

Hi H,

Thanks for the question! This is a really tricky one in my opinion, so I'm glad to have a chance to explain it in more detail.

The conditional rule is triggered by the sufficient condition of C ----- B. So, we could say that the first possibility is one where C ----- B occurs. When that happens, we know B ----- A, leading to: C ----- B ----- A. That is the only solution where C ----- B, and it accounts for one of the four possibilities. But what occurs when C ----- B isn't met, and instead it is B ----- C? That's where the other three possibilities come in:

• When B is first in the chain, and C is second in the chain: B ----- C ----- A.
  
  When B is first in the chain, and C is third in the chain: B ----- A ----- C.
  
  When B is second in the chain, and C is third in the chain: A ----- B ----- C (this is the contrapositive of the conditional rule).

Why do these seemingly "extra" orders occur? Because there are three variables in play, not two. When the conditions are just a single variable each (such as A B), there are only two possibilities for each variable. But when each condition is a chain with two variables, that opens up more possible orders.

Please let me know if that helps. Thanks!