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Six mobsters

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by GmatKiss » Thu May 24, 2012 10:31 pm
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 Anurag@Gurome » Thu May 24, 2012 10:38 pm
GmatKiss wrote: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

No. of ways to arrange six mobsters = 6! = 6 * 5 * 4 * 3 * 2 = 720 ways
In 50% of 720 ways = 360 ways, Frankie will be behind Joey, and in 360 ways Joey will be behind Frankie.

The correct answer is D.
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by GmatKiss » Thu May 24, 2012 11:24 pm
In 50% of 720 ways = 360 ways, Frankie will be behind Joey, and in 360 ways Joey will be behind Frankie

How is this determined!
Please help me understand.
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by GMATGuruNY » Fri May 25, 2012 2:39 am
GmatKiss wrote:In 50% of 720 ways = 360 ways, Frankie will be behind Joey, and in 360 ways Joey will be behind Frankie

How is this determined!
Please help me understand.
I posted an explanation here:

https://www.beatthegmat.com/combinatoric ... 73855.html
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