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Six mobsters have arrived at the theater for

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sharad
 Post subject: Six mobsters have arrived at the theater for
 Post Posted: Sat Sep 29, 2007 4:05 pm 
Q from MGMAT CAT#3:

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?

Ans Choices:
a. 6
b. 24
c. 120
d. 360
e. 720
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sharad
 Post subject: Get a different ans...
 Post Posted: Sat Sep 29, 2007 4:10 pm 
Treating Joey and Franky as one person as they need to be together we have 5! ways to arrange the 6 mobsters...
Then accounting for Joey or Franky to be ahead or behind each other... though the ans should be: 2x5! = 240.

Above ans is not among the ans choices... :shock: the given correct ans is 360 - and the explanation given was not satisfactory...
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StaceyKoprince
 Post subject:
 Post Posted: Tue Oct 02, 2007 5:34 pm 
Offline
ManhattanGMAT Staff


Posts: 6077
Location: San Francisco
Please make sure to use the correct subject header naming convention - the first 8 words of the problem should be your subject header.

Also, you can use the text of the problem to search in case the problem has been posted before, as this one has:
http://www.manhattangmat.com/forums/six-mobsters-have-arrived-at-the-theater-for-the-premiere-of-t739.html?highlight=mobsters

Note that you are assuming something with the wording - you are assuming that Frankie is directly behind Joey. The problem only says that Frankie wants to be behind Joey in general - so Joey could be first in line and Frankie could be last in line, and Frankie would still be behind Joey.

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Stacey Koprince
Instructor
Director of Online Community
ManhattanGMAT
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srikant
 Post subject: you have to use permutation not combination
 Post Posted: Mon Nov 05, 2007 2:06 am 
There are 6 places you have to fill with 6 people, which can be done in 6! ways i.e. 6P6 not 6C6.
nPr= n!/(n-r)!
Now Franky and Joey can change places with each other and there are only 2 positions. Either Franky will be ahead of Joey or Franky will be behind Joey.
Therefore if we divide the entire number of possible arrangements by 2 we will get the total no. of ways either Franky will be behind Joey or ahead of Joey.

6!/2=360
Top 
KK
 Post subject: Question Regarding Moster problem
 Post Posted: Thu Aug 28, 2008 2:41 pm 
Hello Stacey

i was going through the problem and according to the Strategy Guide, i thought this was the way to do it.

5* (5/6+4/6+3/6+2/6+1/6) * 4*3*2*1
= 300

what mistake i am making here? please guide
Top 
RonPurewal
 Post subject: Re: Question Regarding Moster problem
 Post Posted: Tue Sep 09, 2008 7:52 am 
Offline
ManhattanGMAT Staff


Posts: 7146
KK wrote:
Hello Stacey

i was going through the problem and according to the Strategy Guide, i thought this was the way to do it.

5* (5/6+4/6+3/6+2/6+1/6) * 4*3*2*1
= 300

what mistake i am making here? please guide


hi -

could you please post a fuller explanation of where you got this expression?

if we have your explanation, we'll be better able to say what went right and what went wrong.

thanks
-- ron
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