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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]

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New post Updated on: 12 Nov 2014, 07:48
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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, though not necessarily right behind him. 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

Originally posted by reply2spg on 01 Nov 2009, 10:47.
Last edited by Bunuel on 12 Nov 2014, 07:48, edited 2 times in total.
Edited the question and added the OA
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Re: Six mobsters have arrived at the theater for the premiere of  [#permalink]

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New post 01 Nov 2009, 10:54
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reply2spg 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, though not necessarily right behind him. How many ways can the six arrange themselves in line such that Frankie’s requirement is satisfied?

6
24
120
360
720

Please explain

OA D


Total arrangement of 6 = 6!. In half of the cases Frankie will be behind Joey and in half of the cases Joey will be behind Frankie (as probability doesn't favor any of them). So, the needed arrangement is 6!/2=360.

Answer: D (360)
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Re: Six mobsters have arrived at the theater for the premiere of  [#permalink]

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New post 01 Nov 2009, 12:23
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for a more complicated way

assume following scenarios where joey stands 1st, 2nd, 3rd, 4, 5 in line

then Frankie can stand on (5 + 4 + 3 + 2 + 1) spots

multiply that by 4! for the other 4 people

15 x 4! = 360
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Re: Six mobsters have arrived at the theater for the premiere of  [#permalink]

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New post 02 Nov 2009, 10:32
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Wayxi wrote:
I understand the point up to 6!/2=320. But don't we also have to take into account "not necessarily right behind him" ??? So there should be a person between Frankie and Joey ..?


This is exactly what has been taken into account. Consider this: no matter how this 6 will be arranged there can be only two scenarios, either Frankie is behind Joey (when saying behind I mean not just right behind but simply behind, there may be any number of persons between them) OR Joey is behind Frankie. Well, this is pretty obvious. So, as 6 can be arranged in 6! ways, so half of this 6! ways will have *F*J* and half *J*F*. 6!/2=360.

Hope it's clear.
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New post 02 Nov 2009, 11:34
Bunuel,

Thanks for your reply. A lot of GMAT questions are tricky in their wording. This can cause me to OVERanalyze a passage.

Heres how i worked it out:

6! = Total possible combinations
So half of, Joey is in front and the other half Frankie is.

6!/2 = 320 = The half where Joey is in front and Frankie is in back.

Then it states:
But that doesn't count "insists upon standing behind Joey in line at the concession stand, though not necessarily right behind him"

Heres my interpretation of that statement:

It can't be x,x,x,x,F,J
Rather it can only be x,x,x,F,x,J

I don't know... I maybe goin crazy !!!
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New post 02 Nov 2009, 11:57
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Wayxi wrote:
Bunuel,

Thanks for your reply. A lot of GMAT questions are tricky in their wording. This can cause me to OVERanalyze a passage.

Heres how i worked it out:

6! = Total possible combinations
So half of, Joey is in front and the other half Frankie is.

6!/2 = 320 = The half where Joey is in front and Frankie is in back.

Then it states:
But that doesn't count "insists upon standing behind Joey in line at the concession stand, though not necessarily right behind him"

Heres my interpretation of that statement:

It can't be x,x,x,x,F,J
Rather it can only be x,x,x,F,x,J

I don't know... I maybe goin crazy !!!


"not necessarily right behind him" means that he may or may not be right behind him. It's not necessary Frankie to be right behind Joey but if this happens still no problem.

So, **F**J good as well as **FJ**.
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New post 14 Nov 2009, 06:50
man you are a genius !!

you look at the problem from a very simple angle..

Bunuel wrote:
reply2spg 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, though not necessarily right behind him. How many ways can the six arrange themselves in line such that Frankie’s requirement is satisfied?

6
24
120
360
720

Please explain

OA D


Arrangement of 6=6!. In half of the cases Frankie will be behind Joey and in half of the cases Joey will be behind Frankie (as probability doesn't favor any of them). So, the needed arrangement is 6!/2=360.

Answer: D (360)
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Re: Six mobsters have arrived at the theater for the premiere of  [#permalink]

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New post 10 Jun 2010, 05:46
Thanks Bunuel. I should have looked for the same question in an earlier post before initiating a new thread. I have now deleted my post, so that people dont get confused. Nice explanation! It is very clear now.
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New post 05 Jul 2010, 13:27
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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: 700-800 Combinatorics hard concept  [#permalink]

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New post 05 Jul 2010, 15:15
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Quote:
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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Six mobsters have arrived at the theater for the premiere of  [#permalink]

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New post 24 Apr 2011, 16:09
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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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New post 11 May 2012, 04:11
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New post 15 May 2012, 08:18
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Another way that I tried to solve..may be lengthier too...

5! + 4(4!) + 3 (4!) + 2 (4!) + 1(4!) = 360

Assume that in the first case: that XXXXXJ - Frankie total number of people can be arranged in 5! ways.. <Frankie will be always behind Joey>

Second case: XXXXJX - Given that F has to behind J..if F is fixed at 4th then remaining can be arranged in 4! ways and this has to repeated 4 times... so 4(4!)..same has to be done when F is in third position and so on...

I hope it helps!
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New post 09 Sep 2012, 21:41
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This question is easily solved using symmetry. Check out this post to understand the principles used:

http://www.veritasprep.com/blog/2011/10 ... s-part-ii/
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New post 11 Dec 2012, 22:43
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Here's how i did it. Hope it helps someone.

Since F has to be behind J.

It would be better if you write them ("JXXXXF") vertically, i am doing horizontally to save space.

JXXXXF JXXXFX JXXFXX JXFXXX JFXXXX

24 Ways 24 Ways 24 Ways 24 Ways 24 Ways = 120 Ways

(24 ways because 4X's could be rearranged in 4! ways)

XJXXXF XJXXFX XJXFXX XJFXXX

24 Ways 24 Ways 24 Ways 24 Ways = 96 Ways

XXJXXF XXJXFX XXJFXX

24 Ways 24 Ways 24 Ways = 72 Ways

XXXJXF XXXJFX

24 Ways 24 Ways = 48 Ways

XXXXJF

24 Ways = 24 Ways


Total = 120 + 96 + 72 + 48+ 24 = 360.
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New post 27 Dec 2012, 23:05
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reply2spg 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, though not necessarily right behind him. 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


Solution 1:
How many ways to arrange 6 without restriction? 6!
In the total of ways, half of the time J will be behind F or F behind J.
\(=6!/2 = 360\)

Solution 2:
To get a little more practice on permutations...

We could count the possibilities of J(...)F.

0 person in between: 5
1 person in between: 4
2 persons in between: 3
3 persons in between: 2
4 persons in between: 1

(5 + 4 + 3 + 2 + 1) * 4! = 360
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New post 31 Aug 2015, 21:38
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Now the method I'm discussing has already been shared, but the explanations were not too clear for me, so sharing a detailed explanation.
Now following can be the available arrangements (please note that the question requires us to place Frankie behind Joey in line, though not necessarily right behind him):
_ _ _ _ _ F: when F is at the last, there are 5 options for J and the rest can be arranged in 4! ways: 5x4!
_ _ _ _ F _: when F is at the second last position, there are 4 options for J and the rest can be arranged in 4! ways: 4x4!
_ _ _ F _ _: when F is at the third last position, there are 3 options for J and the rest can be arranged in 4! ways: 3x4!
_ _ F _ _ _: when F is at the third third spot from the front, there are 2 options for J and the rest can be arranged in 4! ways: 2x4!
_ F _ _ _: when F is at the third third spot from the front, there is only 1 options for J and the rest can be arranged in 4! ways: 4!
Now F can not be placed at the first spot as he has to be behind J, so above are al the situations possible and the sum would give us the total arrangements.
total arrangements = 5x4! + 4x4! + 3x4! + 2x4! + 4! = (5+4+3+2+1)4! = 15x4! = 360
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New post 01 Sep 2015, 03:27
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, though not necessarily right behind him. 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
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New post 12 Aug 2018, 22:04
Bunuel

My take -

Total 6 places to fill.

_ _ _ _ _ J. ( No. Of place for F is 5 OR F can take any place among the 5)

_ _ _ _ J _ .( Places for F = 4)
Similarly
F will have 3,2,1 places as per J position.
Total ways = 5*4*3*2*1=120.

Where did I go wrong? Please explain !
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New post 13 Aug 2018, 10:06
AdityaHongunti wrote:
Bunuel

My take -

Total 6 places to fill.

_ _ _ _ _ J. ( No. Of place for F is 5 OR F can take any place among the 5)

_ _ _ _ J _ .( Places for F = 4)
Similarly
F will have 3,2,1 places as per J position.
Total ways = 5*4*3*2*1=120.

Where did I go wrong? Please explain !
chetan2u

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You are not taking the combinations in which others can stand...
Say I wanted to do as per your way..
1) _,_,_,_,_, J
So remaining 5 in 5!
2) _,_,_,_, J,_
F can take any of 4, and the remaining 4 seats can be taken in 4! Ways
So 4*4!
3) _,_,_, J,_,_
Similarly 3*4!
4) _,_, J,_,_,_
So 2*4!
5) _, J,_,_,_,_
So 1*4!

Total
5!+4*4!+3*4!+2*4!+1*4!=4!(5+4+3+2+1)=24*15=360..

Hope it helps
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Re: Six mobsters have arrived at the theater for the premiere of   [#permalink] 13 Aug 2018, 10:06

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