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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 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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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
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01 Nov 2009, 10:54
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 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
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01 Nov 2009, 12:23
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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02 Nov 2009, 10:32



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Re: Six mobsters have arrived at the theater for the premiere of
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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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Re: Six mobsters have arrived at the theater for the premiere of
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02 Nov 2009, 11:57
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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Re: Six mobsters have arrived at the theater for the premiere of
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02 Nov 2009, 12:08
Bunuel,
Ahhhhh, got it, thanks a lot.



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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 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
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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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Six mobsters have arrived at the theater for the premiere of
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22 Aug 2010, 10:46
hi, 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? if i use the Glue method introduced in MGMAT book, i glue Frankie and Joey together, i get 5!=120. please advise why my answer is wrong. thanks! bunuel, what's wrong with my glue method??



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Re: Six mobsters have arrived at the theater for the premiere of
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22 Aug 2010, 11:58
tt11234 wrote: bunuel, what's wrong with my glue method?? This issue is discussed in previous posts: 5!=120 is # of ways to arrange Frankie and Joey so that Frankie is right behind Joey, but it's not necessary Frankie to be right behind Joey, there can be 0, 1, 2, 3, or 4 people between them. Hope it's clear.
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Six mobsters have arrived at the theater for the premiere of
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11 May 2012, 04:11



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15 May 2012, 08:18
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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Re: Six mobsters have arrived at the theater for the premiere of
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09 Sep 2012, 11:47
Bunuel My question is in the problem i is stated as "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." J vvill be behind F but F beind J is not possible please clarify 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 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
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09 Sep 2012, 21:41
This question is easily solved using symmetry. Check out this post to understand the principles used: http://www.veritasprep.com/blog/2011/10 ... spartii/
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Re: Six mobsters have arrived at the theater for the premiere of
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11 Sep 2012, 03:03
venmic wrote: Bunuel My question is in the problem i is stated as "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." J vvill be behind F but F beind J is not possibleplease clarify 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 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) We need all the cases where F is behind J and since the number of all arrangements is 6!, then the cases when F is behind J is 6!/2.
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Re: Six mobsters have arrived at the theater for the premiere of
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11 Dec 2012, 22:43
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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Re: Six mobsters have arrived at the theater for the premiere of
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26 Dec 2012, 02:40
Bunuel wrote: piyushksharma 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 Merging similar topics. Questions about the same concept to practice: susanjohndaisytimmattandkimneedtobeseatedin130743.htmlmegandbobareamongthe5participantsinacyclingrace58095.htmlinhowmanydifferentwayscanthelettersaab91460.htmlmaryandjoearetothrowthreediceeachthescoreisthe126407.htmlgoldenrodandnohopeareinahorseracewith6contestants82214.htmlHope it helps. Good reference problems! 2 questions tho: 1. If the question says right F is right behind J, then isn't it the same way to solve when we divide 6! by 2? Because either F is right behind J or F is not, so we have 2 cases. This is wrong according to OA, so how to correctly recognize the cases on which the probability favors? 2. Extension: can we apply the "parity" concept (yes/no, yin/yang, with/without, positive/negative etc) in problems that take 3 (or more) cases neither of which the probability favors? So I can just calculate the total number of way and then divide by 3 for the solution? Thanks!



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Re: Six mobsters have arrived at the theater for the premiere of
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26 Dec 2012, 03:06
catennacio wrote: Bunuel wrote: piyushksharma 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 Merging similar topics. Questions about the same concept to practice: susanjohndaisytimmattandkimneedtobeseatedin130743.htmlmegandbobareamongthe5participantsinacyclingrace58095.htmlinhowmanydifferentwayscanthelettersaab91460.htmlmaryandjoearetothrowthreediceeachthescoreisthe126407.htmlgoldenrodandnohopeareinahorseracewith6contestants82214.htmlHope it helps. Good reference problems! 2 questions tho: 1. If the question says right F is right behind J, then isn't it the same way to solve when we divide 6! by 2? Because either F is right behind J or F is not, so we have 2 cases. This is wrong according to OA, so how to correctly recognize the cases on which the probability favors? 2. Extension: can we apply the "parity" concept (yes/no, yin/yang, with/without, positive/negative etc) in problems that take 3 (or more) cases neither of which the probability favors? So I can just calculate the total number of way and then divide by 3 for the solution? Thanks! 6!/2 gives the cases when F is not right behind J, for example, *J*F***.
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Re: Six mobsters have arrived at the theater for the premiere of
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27 Dec 2012, 23:05
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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