Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

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.
_________________

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?

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

6 people can be arranged in 6! ways (= 720). Frankie will be either ahead of Joey or behind him in each one of these cases. In half of these, Frankie will be ahead of Joey and in half, he will be behind.

Therefore, number of cases in which Frankie is behind Joey is 720/2 = 360

the total number of options: \(P^6_6=6!=720\) All options are divided by two symmetry possibilities: Frankie behind Joey or Joey behind Frankie Therefore, \(N=\frac{720}{2}=360\)

Considering all the options(6!) there can be only two ways Frankie and Joey can be: Frankie behind Joey or Joey behind Frankie. To get one possibility divide by 2. Hope this

Re: Arrangements(Combinatorics): Six mobsters for Goodbuddies [#permalink]

Show Tags

12 Oct 2011, 13:12

The questions says that frank is behind george, that in no way means they are together, next to each other.

the cases would be :-

1. G||||| : F could be anywhere behind G, so 5! ways + 2.XG|||| : F could be any one of |, so 4* 4! ways + 3.XXG||| : F could be any one of |, so 4*3*3! ways 4.XXXG|| : F could be any one of |, so 4*3*2*2! ways 5.XXXXG| : F could be just |, so 4*3*2*1*1! ways

adding 1,2,3,4,5, it comesa to 120+96+72+48+24 = 360. answer option D.
_________________

Paras.

If you found my post helpful give KUDOS!!! Everytime you thank me but don't give Kudos, an Angel dies!

I am now providing personalized one to one GMAT coaching over Skype at a nominal fee. Hurry up to get an early bird discount! Send me an IM to know more.

gmatclubot

Re: Arrangements(Combinatorics): Six mobsters for Goodbuddies
[#permalink]
12 Oct 2011, 13:12

Happy New Year everyone! Before I get started on this post, and well, restarted on this blog in general, I wanted to mention something. For the past several months...

It’s quickly approaching two years since I last wrote anything on this blog. A lot has happened since then. When I last posted, I had just gotten back from...

Post-MBA I became very intrigued by how senior leaders navigated their career progression. It was also at this time that I realized I learned nothing about this during my...