MGMAT CAT # 1 (combination)

[scared-stiff]

Question from Manhattan GMAT CAT # 1

Six mobsters have arrived at the theater for the premiere of the film "Goodbuddies." One of the mobsters, Frankie, is an informer, and he's afraid that another member of his crew, Joey, is on to him. Frankie, wanting to keep Joey in his sights, insists upon standing behind Joey in line at the concession stand. How many ways can the six arrange themselves in line such that Frankie’s requirement is satisfied?

a. 6
b. 24
c. 120
d. 360
e. 720

( Official answer is d. and I am not buying it)

[shaji]

The official answer is correct.

There are two and only two options here; ie Frankie is behind or ahead of Joe. The ways these 'guys can stand is 6!(720) ways. So there are 360(720/2) ways that Frankie is in Joe's sights.

Question from Manhattan GMAT CAT # 1

Six mobsters have arrived at the theater for the premiere of the film "Goodbuddies." One of the mobsters, Frankie, is an informer, and he's afraid that another member of his crew, Joey, is on to him. Frankie, wanting to keep Joey in his sights, insists upon standing behind Joey in line at the concession stand. How many ways can the six arrange themselves in line such that Frankie’s requirement is satisfied?

a. 6
b. 24
c. 120
d. 360
e. 720

( Official answer is d. and I am not buying it)

[scared-stiff]

[quote="shaji"]The official answer is correct.

There are two and only two options here; ie Frankie is behind or ahead of Joe. The ways these 'guys can stand is 6!(720) ways. So there are 360(720/2) ways that Frankie is in Joe's sights.

[quote="scared-stiff"]Question from Manhattan GMAT CAT # 1

the question says
"Frankie, wanting to keep Joey in his sights, insists upon standing behind Joey in line at the concession stand"

So why is Frankie ahead of Joe ? Also, does behind mean directly behind (J <- F) or just behind (J <- x <- x <-x F)

What you wrote is the official explanation and I am not getting it. (thanks for replying though)

[shaji]

Frankie has to be behind Joe to keep him in sights. Otherwise Joe will have Frankie in his sights which MUST be avoided.

The official approach is indeed the quickest approach.

Now!, to tickle your mind, consider the number of ways they can stand if Frankie has to be behind Joe and Joe behind Pat.!!!

The official answer is correct.

There are two and only two options here; ie Frankie is behind or ahead of Joe. The ways these 'guys can stand is 6!(720) ways. So there are 360(720/2) ways that Frankie is in Joe's sights.

Question from Manhattan GMAT CAT # 1

the question says
"Frankie, wanting to keep Joey in his sights, insists upon standing behind Joey in line at the concession stand"

So why is Frankie ahead of Joe ? Also, does behind mean directly behind (J <- F) or just behind (J <- x <- x <-x F)

What you wrote is the official explanation and I am not getting it. (thanks for replying though)

[Guest]

Maybe the confusion is caused by the original poster reading the problem as "Frankie must stand directly behind Joey in line". In this case, you essentially reduce the two mobsters to one unit and the answer would be 5!=120.

However, this is not what the problem stated. Frankie must be behind Joey, but not necessarily directly behind him. Without any restrictions the six mobsters can line up in 6! = 720 ways. In half these 720 instances Frankie will be behind Joey, and in front of him in the other half. So the answer is 720/2 = 360.

[scared stiff]

In half these 720 instances Frankie will be behind Joey, and in front of him in the other half. So the answer is 720/2 = 360.

WHY ? This is what I am trying to understand. If this is permutation concept then let please me know.
I am trying to understand the quick way !6/2.

I finally solved it the *LONG* way (at least I can understand the issue)

Here's how I think

J _ _ _ _ _ == !5 ways for F to stand = 120
_ J _ _ _ _ == 4 . !4 ...... = 96
_ _ J _ _ _ == 4.3.!3 = 72
_ _ _ J _ _ == 4.3.2.!2 = 48
_ _ _ _ J _ == 4.3.2.1.!1 = 24

TOTAL 360

[Guest]

Scared stiff -

Solving it the long way to get to 360 is entirely correct and probably not that time consuming. Here's the argument that let's you just chop 720 in half:

1) Joey standing in front of Frankie and Frankie standing in front of Joey are complimentary events (i.e. can't be true at the same time AND one or the other MUST be true).

2) Then just argue from symmetry that the number of permutations where frankie is in front of Joey is equal to the number of ways that joey is in front of frankie. Not sure how to formally prove this, but look at the diagram you drew in your response above. If you filled in an "F" instead of a "J" in the blanks above you'll still come up with 360.

3) From 2) you know that the number of "good" permutations (Joey in front of Frankie) is equal to the number of bad permutations (Frankie in front of Joey). From 1) you know that these account for all 720 possible permutations, so the total number of "good" permutations = 720/2 = 360.

[RonPurewal]

Re: last 'guest' post:

In point 2 you mentioned formal proof. Although it's utterly irrelevant to the GMAT, a formal proof would involve a bijection (1-1 correspondence) between the orders in which F is ahead of J and those in which J is ahead of F. Each arrangement with F in front of J in uniquely matched to an arrangement with J in front of F (and vice versa), merely by switching F and J. This establishes the desired 1-1 correspondence, so each set contains exactly half of the possibilities.

Other than that, you've got everything covered pretty well.

The wording of this problem is admittedly more confusing than it could be, because of the common meaning of 'behind': "the guy behind Brad" would unambiguously be taken by just about everyone to refer to the guy _directly_ behind Brad. On the actual exam, this problem would most likely be worded more clearly, like 'Joey is closer to the front of the line than is Frankie.'