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Six mobsters have arrived at the theater for the premiere of

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Six mobsters have arrived at the theater for the premiere of [#permalink] New post 24 Jul 2007, 06:08
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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

OPEN DISCUSSION OF THIS QUESTION IS HERE: six-mobsters-have-arrived-at-the-theater-for-the-premiere-of-the-126151.html
[Reveal] Spoiler: OA
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Re: PS: Combinatorics [#permalink] New post 18 Mar 2008, 21:28
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360

If Joey is in 6th spot, Frank can stand anywhere from 1-5 - rest can be arranged in 4! ways - total 5*4!
If Joey is in 5th spot, Frank can stand anywhere from 1-4 - rest can be arranged in 4! ways - total 4*4!
.
.
Total number of arrangments = (1+2+3+4+5)*4! = 360
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 [#permalink] New post 02 Dec 2007, 09:38
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I think this is the clearest way to look at it:

JF1234
12JF34
1234JF

Since JF is considered 1 group, the total number of combinations would be 5!. Since there are 3 ways in which JF can be together, we multiply 5! by 3, therefore:

5!*3= 360

hope that's clearer now
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Re: Mobsters going for a movie [#permalink] New post 13 Sep 2008, 18:14
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Here's how I did it, but it seems long and cumbersome so hopefully someone can provide a better explanation.

There are 6 possible positions for each mobster. Frankie depends on Joey so select those first

1. Joey at 1; 5! for rest
2. Joey at 2; 4 choices for 1 (6-joey-frankie); 4! for rest
3. Joey at 3; 4*3 choices for 1-2; 3! for rest
4. Joey at 4; 4*3*2 choices for 1-3; 2! for rest
5. Joey at 5; 4*3*2*1 choice for 1-4; 1 for rest

So total = 5! + 4x4! + 4*3*3! + 4*3*2*2! + 4!
= 5(4!) + 4(4!) + 3(4!) + 2(4!) + (4!)
= 4! x (5 + 4 + 3 + 2 + 1)
= 24 * 15
= 360


So D.
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Re: 700-800 Combinatorics hard concept [#permalink] New post 05 Jul 2010, 12:27
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Here's the answer.

There are fifteen ways F would be behind J all the time. Think of it this way. There are six places, and you have to select two places for F & J, and where the other four's positions does not matter. This can be put into numbers as 6!/4!2! giving us 15.

So you'd have the following:

FJ????, F?J???, F??J??, F???J?, F????J
?FJ???, ?F?J??, ?F??J?, ?F???J
??FJ??, ??F?J?, ??F??J
???FJ?, ???F?J
????FJ

15 scenarios in all.

Now, you can use counting techniques to get all the possible orders of the four remaining mobsters. That would be 4! = 24.

Multiply the 15 cases with all of the possible orders of the four remaining mobsters - 15*24 = 360.
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Re: Combination [#permalink] New post 24 Nov 2007, 12:54
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tarek99 wrote:
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

I really don't know how to approach this problem. Would anyone show me how? appreciate it!


1. If JF are adjacent to each other = 5! = 120
1. If F is behind but not adjcent to J = 3 x 5! = 360
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Re: 700-800 Combinatorics hard concept [#permalink] New post 05 Jul 2010, 14:15
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I have a hard time grasping the concept that half of the time F would be behind J. Can someone please explain this to me?


Hi,

maybe this will help:

take 2 objects, x and y. How many ways to arrange them?

Clearly, 2: xy and yx. Notice that in half the arrangements (1 out of 2) x is before y, and in the other half y is before x.

Take 3 objects, x, w, and y. There are 3! or 6 arrangements. In half (ie, 3) of those arrangements, x is before w, while in the other half w before x. Likewise, in half of the arrangements x is before y while in the other half y before x. And, also, in half the arrangements, w is before y while in the other half y before w.

Why would it be the case that Joey can be arranged ahead of Frankie more or less often than Frankie can be arranged ahead of Joey? Why not the other way around? :wink:

So, here, the easiest way to solve is certainly to take just half of the total arrangements: 6!/2 = 360, and choose D.

Quote:
Also, can this be down using straight combinations without getting the concept that half of the time F would be behind J?


Yes, it certainly can as the above poster demonstrated!

But many combinatorics questions on the GMAT resist pure formulaic treatment. A little bit of reasoning on these questions can save you an immense amount of time!
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Re: Permutation & Combination [#permalink] New post 24 Apr 2011, 15:09
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dumluck wrote:
Sorry to drag this post up again, but rather than start a new one. I understand that 6! is the number of combinations. However, the question explicitly states that Frankie has to be behind Joey (not directly but certainely behind). So why are we dividing by 2? As in; the probability that Frankie will be behind Joey and the probability that Frankie will be ahead of Joey if we are only concerned with the former (i.e Frankie behind Joey)?

Apologies, but I'm just not getting it.


The important thing to realize here is that Frankie and Joey are absolutely identical elements of this arrangement.
Say, I have 3 elements: A, B and C
I can arrange them in 3! ways:

ABC
ACB
BAC
BCA
CAB
CBA
Look at them carefully. In 3 of them A is before B and in other 3, B is before A. It will be this way because A and B are equal elements. There is no reason why A should be before B in more cases than B before A. Similarly, you can compare B and C or A and C.
Hence, when we arrange all 6 people in 6! ways, in half of them Frankie will be before Joey and in other half, Joey will be before Frankie.
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Re: PS: Mobsters [#permalink] New post 24 Jul 2007, 06:13
trahul4 wrote:
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


Since F and J are to be always together, number of arrangements is 5! (considering JF as one). Hence, 120 is the answer.

Here is a discussion thread for the same problem:
http://www.gmatclub.com/phpbb/viewtopic.php?t=45097&highlight=mobster
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 [#permalink] New post 24 Jul 2007, 06:23
Since Frankie and Joey must always be in a certain order, we can take them as one entity.

So the number of ways to arrange them = 5! = 120
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 [#permalink] New post 24 Jul 2007, 06:44
Yes looks very easy problem.
But OA is 360. I got this question in one of the MGMAT CAT and was surprised to see 120 worng. Anyways here is the explaination given by MGMAT.

Ignoring Frankie's requirement for a moment, observe that the six mobsters can be arranged 6! or 6 x 5 x 4 x 3 x 2 x 1 = 720 different ways in the concession stand line. In each of those 720 arrangements, Frankie must be either ahead of or behind Joey. Logically, since the combinations favor neither Frankie nor Joey, each would be behind the other in precisely half of the arrangements. Therefore, in order to satisfy Frankie's requirement, the six mobsters could be arranged in 720/2 = 360 different ways.

The correct answer is D.

I feel this soultion given by MGMAT is incorrect.
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 [#permalink] New post 24 Jul 2007, 07:02
trahul4 wrote:
Yes looks very easy problem.
But OA is 360. I got this question in one of the MGMAT CAT and was surprised to see 120 worng. Anyways here is the explaination given by MGMAT.

Ignoring Frankie's requirement for a moment, observe that the six mobsters can be arranged 6! or 6 x 5 x 4 x 3 x 2 x 1 = 720 different ways in the concession stand line. In each of those 720 arrangements, Frankie must be either ahead of or behind Joey. Logically, since the combinations favor neither Frankie nor Joey, each would be behind the other in precisely half of the arrangements. Therefore, in order to satisfy Frankie's requirement, the six mobsters could be arranged in 720/2 = 360 different ways.

The correct answer is D.

I feel this soultion given by MGMAT is incorrect.


Sounds dodgy.. okay, let's try the long-winded way. We know Frankie always wants to stand behind Joey.

So,
- if Joey is first in the queue, Frankie must be second. The remaining 4 mobsters can be arranged 4! ways.

- if Joey is second in the queue, Frankie must be third. The remaining 4 mobsters can be arranged 4! ways.

- if Joey is third in the queue, Frankie must be fourth. The remaining 4 mobsters can be arranged 4! ways.

- if Joey is fourth in the queue, Frankie must be fifth. The remaining 4 mobsters can be arranged 4! ways.

- if Joey is fifth in the queue, Frankie must be last. The remaining 4 mobsters can be arranged 4! ways.

So total = 5 * 4! = 120

This method should be foolproof as we are just shifting the pair along the queue. Don't see how you can get 360.
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 [#permalink] New post 24 Jul 2007, 07:13
I agree with you. I think the OA given by MGMAT is wrong and my answer on the CAT was correct. :wink:
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 [#permalink] New post 24 Jul 2007, 07:48
trahul4 wrote:
I agree with you. I think the OA given by MGMAT is wrong and my answer on the CAT was correct. :wink:


I had the same problem with MGMAT. This problem has been discussed several times on this board. MGMAT is actually asking how many ways for Frankie to be behind Joey but not directly behind him. Thus, the ways of arranging the 6 people is 6!=720 and Frankie is behind Joey 1/2 of the times so the answer is 360. The answer allows for there to be mobsters between Joey and Frankie.

MGMAT was not clear on this question in my opinion. However, since they did not specify "directly behind" then I guess their answer is okay.
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 [#permalink] New post 26 Jul 2007, 18:06
I hope the wording isn't this tricky on the actual test! I also got 120 until I read the explanation, assuming behind meant directly behind.
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Combination [#permalink] New post 24 Nov 2007, 12:40
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

I really don't know how to approach this problem. Would anyone show me how? appreciate it!
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 [#permalink] New post 24 Nov 2007, 13:06
thank you and the OA is 360. here is how i approached it. because J and F must be next to each other, the whole arrangement is 5! then, because they have to be next to each other, i also considered the 2!. so when i multiple 5!*2!, i get 240. yet this approach is wrong. i don't understand why you used just 3. why is it wrong to use the 2! as well? i understand the 5!, but don't understand why you used just 3 instead of the 2!

would you explain? thanks
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 [#permalink] New post 24 Nov 2007, 15:35
tarek99 wrote:
thank you and the OA is 360. here is how i approached it. because J and F must be next to each other, the whole arrangement is 5! then, because they have to be next to each other, i also considered the 2!. so when i multiple 5!*2!, i get 240. yet this approach is wrong. i don't understand why you used just 3. why is it wrong to use the 2! as well? i understand the 5!, but don't understand why you used just 3 instead of the 2!

would you explain? thanks



= 5 x 4! + 4 x 4! + 3 x 4! + 2 x 4! + 4!
= 4! (5+4+3 +2 + 1)
= 4! 15
= 4! x 5x3
= 5! x 3
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 [#permalink] New post 24 Nov 2007, 18:12
D.

(5!)(15/5) = (120)(3)=360

Where 5! represents the ways in which the mobsters-JF can sit. 15/5 are the ways in which JF can sit.
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 [#permalink] New post 25 Nov 2007, 01:10
spider wrote:
GMAT TIGER wrote:
tarek99 wrote:
thank you and the OA is 360. here is how i approached it. because J and F must be next to each other, the whole arrangement is 5! then, because they have to be next to each other, i also considered the 2!. so when i multiple 5!*2!, i get 240. yet this approach is wrong. i don't understand why you used just 3. why is it wrong to use the 2! as well? i understand the 5!, but don't understand why you used just 3 instead of the 2!

would you explain? thanks



= 5 x 4! + 4 x 4! + 3 x 4! + 2 x 4! + 4!
= 4! (5+4+3 +2 + 1)
= 4! 15
= 4! x 5x3
= 5! x 3


I dint get this? anyone please explain!


ok its bit tricky:

1. JF1234 = 5 x 4!
2. 1JF234 = 4 x 4!
3. 12JF34 = 3 x 4!
4. 123JF4 = 2 x 4!
5. 1234JF = 1 x4!

add them up = 5 x 4! + 4 x 4! + 3 x 4! + 2 x 4! + 4! = 5! x 3


hope it is clear now.
  [#permalink] 25 Nov 2007, 01:10
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