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I have all of my conditionals down with contrapositives in my head. And numerical distribution in mind.

Numerical Distribution:

1-1-1-3
1-1-1-1-2

At most one person left more than one message (as shown with distribution above).

No person left more than three messages (as shown with distribution above).

Constraints:

H1 ---> P6

G ---> F and P

F ---> All of P>T

P ---> All of H>L

Question #7

The first and last messages on the answering machine could be the first and second messages left by which one of the following?

A) F
B) H <---- I can discard because if H is in 1 then P is in 6th, thus this cannot be the answer.
C) L
D) P
E) T

This is the only question that I had trouble with. It was nice to have a must be true EXCEPT question in this game, but I do not want to use that in reference to this question (if it is even really applicable at all in this case) as I want to know how to approach this question.

I could only get rid of one answer and my previous work of hypotheticals from other questions did not help on this question.

Let's try F:

F P T H L F

F F P T H L

That works.

Let's try L:

L ...I know that there is no P due to the contrapositive, and then I know that there is no F due to the contrapositive, and I know there is no G due to the contrapositive. I am left with 3 variables and no longer have a valid numerical distribution to handle such a case. Discard.

Let's try P:

P H>L...thats all I have now. I am unsure of what to make of this situation. I am keeping it for now.

Let's try T:

T....I know I cannot have an F because if I did then P would have to become before it, thus T would not be able to go first.

So now I know that I do not have a G because of the contrapositive of the second conditional rule. Without two of the variables, I am left with four to work with. This triggers the 1-1-1-3 distribution. This triggers T going 3 times.

T T P H L T

That works.


Stuck with two answers and I am not seeing how either one is invalid. Please help.

secretad wrote: Numerical Distribution:

1-1-1-3
1-1-1-1-2

You've neglected the distribution in which each of the six leaves one message. For instance, the following is a valid solution:
F P T H L G

secretad wrote: Let's try F:

F P T H L F

F F P T H L

That works.

The second solution is not relevant to this particular question, since F is not assigned to the sixth position.

secretad wrote: Let's try P:

P H>L...thats all I have now. I am unsure of what to make of this situation. I am keeping it for now.

Let's examine this choice. Here's what we have so far:
P1 - H - L - P6

Since T can be assigned to a maximum of one of the remaining two positions (first condition), at least one of F and G must be selected. Either way, the fifth condition will be triggered and violated.

secretad wrote: T T P H L T

That works.

This solution does not conform to the question; to test it, you must only assign T to the first and sixth positions. Removing T from the second position forces either of F and G to be selected, and either way, the fifth condition will be triggered and violated.

How is the t scenario invalid to answering the question. It can be first and last, and it can be first and second. It does not say that you can only put t in the first and last spots. I have t in those spots without violating a rule and I can have t go first and second without violating any rules.

Although it doesn't violate any rules, it doesn't apply to this question. The wording is a bit tricky, but you're looking for a variable which can be assigned to both the first and the sixth position under one viable scenario. Since these messages are the first and second messages left by one particular person, the distribution is fixed at 2, 1, 1, 1, 1. The same two messages are being referenced in two different ways:

• "The first and last messages": 1 2 3 4 5 6
• "the first and second messages left by": the numbers here apply to the variable in question only, not the positions themselves