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#1 (permalink) |
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Eager!
Join Date: Mar 2009
Posts: 61
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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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#5 (permalink) |
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Eager!
Join Date: Mar 2009
Posts: 57
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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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#6 (permalink) |
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I JUST got here.
Join Date: Sep 2009
Posts: 24
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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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#7 (permalink) |
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GMAT TEST EXPERTS
![]() ![]() ![]() Join Date: Nov 2008
Posts: 503
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Yes, i got 6!/2.
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