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Six mobsters have arrived at the theater for the premiere of [#permalink]
01 Nov 2009, 09:47

2

This post was BOOKMARKED

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

Last edited by Bunuel on 15 Apr 2012, 23:50, edited 1 time in total.

Re: Permutation & Combination [#permalink]
01 Nov 2009, 09:54

2

This post received KUDOS

Expert's post

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?

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.

Re: Permutation & Combination [#permalink]
02 Nov 2009, 09:32

Expert's post

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.

Re: Permutation & Combination [#permalink]
02 Nov 2009, 10:57

1

This post received KUDOS

Expert's post

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**. _________________

Re: Permutation & Combination [#permalink]
14 Nov 2009, 05:50

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?

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.

Re: Permutation & Combination [#permalink]
10 Jun 2010, 04:46

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.

glue method from mgmat [#permalink]
22 Aug 2010, 09:46

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!

Re: Permutation & Combination [#permalink]
22 Aug 2010, 10:58

Expert's post

1

This post was BOOKMARKED

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.

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?

Re: Six mobsters have arrived at the theater for the premiere of [#permalink]
15 May 2012, 07:18

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...

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?

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.

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?

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. _________________

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?

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?

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?

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***. _________________