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# Six mobsters have arrived at the theater for the premiere of

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

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Updated on: 05 Oct 2011, 22:24
This is the way I could see the solution clearer:

Total arrangements of 6 mobsters 6! = 760

If we picture all the possible arrangements of lines we will get a rectangle 6 X 120 = 720

J F 1 2 3 4
J 1 F 2 3 4
J 1 2 F 3 4
J 1 2 3 F 4
J 1 2 3 4 F

1 J F 2 3 4
1 J 2 F 3 4

And so on, we will realize that the arrangements will follow a path similar to

1 1 1 1 1
0 1 1 1 1
0 0 1 1 1
0 0 0 1 1
0 0 0 0 1

Clearly half of the total arrangements or 360. Hope it can help some of you!!

Regards

Originally posted by Bull78 on 03 Oct 2011, 18:32.
Last edited by Bull78 on 05 Oct 2011, 22:24, edited 1 time in total.
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Re: PS: Combinatorics  [#permalink]

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03 Oct 2011, 19:04
walker wrote:
another way:

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

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12 Oct 2011, 14:12
1
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: Six mobsters have arrived at the theater for the premiere of  [#permalink]

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31 Aug 2018, 21:06
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Re: Six mobsters have arrived at the theater for the premiere of   [#permalink] 31 Aug 2018, 21:06

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# Six mobsters have arrived at the theater for the premiere of

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