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Sorry to drag this post up again, but rather than start a new one. I understand that 6! is the number of combinations. However, the question explicitly states that Frankie has to be behind Joey (not directly but certainely behind). So why are we dividing by 2? As in; the probability that Frankie will be behind Joey and the probability that Frankie will be ahead of Joey if we are only concerned with the former (i.e Frankie behind Joey)?

Apologies, but I'm just not getting it.

The important thing to realize here is that Frankie and Joey are absolutely identical elements of this arrangement. Say, I have 3 elements: A, B and C I can arrange them in 3! ways:

ABC ACB BAC BCA CAB CBA Look at them carefully. In 3 of them A is before B and in other 3, B is before A. It will be this way because A and B are equal elements. There is no reason why A should be before B in more cases than B before A. Similarly, you can compare B and C or A and C. Hence, when we arrange all 6 people in 6! ways, in half of them Frankie will be before Joey and in other half, Joey will be before Frankie.
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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?

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

6 people can be arranged in 6! ways (= 720). Frankie will be either ahead of Joey or behind him in each one of these cases. In half of these, Frankie will be ahead of Joey and in half, he will be behind.

Therefore, number of cases in which Frankie is behind Joey is 720/2 = 360

the total number of options: \(P^6_6=6!=720\) All options are divided by two symmetry possibilities: Frankie behind Joey or Joey behind Frankie Therefore, \(N=\frac{720}{2}=360\)

Considering all the options(6!) there can be only two ways Frankie and Joey can be: Frankie behind Joey or Joey behind Frankie. To get one possibility divide by 2. Hope this

Re: Arrangements(Combinatorics): Six mobsters for Goodbuddies [#permalink]

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12 Oct 2011, 14:12

The questions says that frank is behind george, that in no way means they are together, next to each other.

the cases would be :-

1. G||||| : F could be anywhere behind G, so 5! ways + 2.XG|||| : F could be any one of |, so 4* 4! ways + 3.XXG||| : F could be any one of |, so 4*3*3! ways 4.XXXG|| : F could be any one of |, so 4*3*2*2! ways 5.XXXXG| : F could be just |, so 4*3*2*1*1! ways

adding 1,2,3,4,5, it comesa to 120+96+72+48+24 = 360. answer option D.
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Re: Arrangements(Combinatorics): Six mobsters for Goodbuddies
[#permalink]
12 Oct 2011, 14:12

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