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Mobsters

by ricky » Tue Aug 05, 2008 4:59 am
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
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by sudhir3127 » Tue Aug 05, 2008 5:04 am
i would say 1/2*6! which is 360.

do let us know the OA.
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by ricky » Tue Aug 05, 2008 5:05 am
It is D..But how?Thnx
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by sudhir3127 » Tue Aug 05, 2008 5:15 am
here it goes...

6 people can arrange in 6! ways = 720 ways.
now

there's a 50% chance Frankie will be ahead of Joey in line, and a 50% chance Frankie will be behind Joey in line hence its

1/2*6! = 360.

hope it helps..
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by ricky » Tue Aug 05, 2008 5:21 am
Thanx Sudhir....got it...
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