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Old 2009 June 8th, 10:26 PM   #1 (permalink)
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Six Mobsters

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?
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Old 2009 June 9th, 02:31 AM   #2 (permalink)
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There are 6 ! ways to arrange the mobsters. and in half of those arrangements J will be behind F. So the answer should be 6!/2.

is there a OA?
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Old 2009 June 9th, 10:26 AM   #3 (permalink)
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buskool,
Great way to solve this..

i took each condition and solved them.. although I got the same answer but it took me 2 mins to solve this..
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Old 2009 June 11th, 06:40 PM   #4 (permalink)
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Quote:
Originally Posted by buskool View Post
There are 6 ! ways to arrange the mobsters. and in half of those arrangements J will be behind F. So the answer should be 6!/2.

is there a OA?
Good approach !
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Old 2009 June 18th, 12:01 PM   #5 (permalink)
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My approach was

Positions 1 2 3 4 5 6
case 1 J F F F F F => 5 different positions possible
2: J F F F F => 4 possible positions
3: J F F F => 3 positions
4: J F F => 2
5: J F => 1
A total of 5+4+3+2+1 = 15 positions possible for just 2 of them
For each of this the rest of them can be arranged in 4! ways (4 positions 4 people)

Hence 15 * 24 = 360
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Old 2009 October 17th, 08:56 AM   #6 (permalink)
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I saw this question on the Manhattan GMAT test forum with the following options

A 6
B 24
C 120
D 360
E 720


another easy way to solve this is through elimination.

6 mobsters... means max possible arrangements are 6! = 720.... this is without any restrictions hence eliminate E

next taking a part of the restrictions...
suppose J & F always stand together, then possible arrangements are:
JF----
-JF---
--JF--
---JF-
----JF
5 possible arrangements for JF and 4! ways to arrange the remaining men..

therefore, total arrangements for this restriction is 5 X 4! = 120...

so this is when we consider only a part of the restrictions, hence 120 cannot be the answer... eliminate C.

therefore we have to get an answer which is greater than 120 and smaller than 360...

We are left with D... voila!
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Old 2009 October 18th, 12:12 PM   #7 (permalink)
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Yes, i got 6!/2.
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http://mbachase.co/index.php?option=com_content&view=article&id=72&It emid=62
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