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Six mobsters have arrived at the theater for the premiere of

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Re: Permutation & Combination [#permalink] New post 24 Apr 2011, 15:09
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dumluck wrote:
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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Re: Permutation & Combination [#permalink] New post 01 May 2011, 01:25
6! / 2 = 360.
6 people can stand in 6! ways.
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Re: Combination [#permalink] New post 02 May 2011, 19:22
6! total arrangements.
half of them will be where F will be at the back of J.

360.
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PS - 700 level - arrangements [#permalink] New post 04 Sep 2011, 13:50
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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Re: PS - 700 level - arrangements [#permalink] New post 04 Sep 2011, 20:06
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bschool83 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


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 theory has been discussed in detail here:
advanced-constraint-combinatorics-42275.html
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Re: Arrangements(Combinatorics): Six mobsters for Goodbuddies [#permalink] New post 14 Sep 2011, 11:38
D : 360 is the right answer

Here it goes

- - - - - -

Six places to be filled ,so we need to make sure J and F meet the criteria such that F is always behind J

so 6C2 will give the possible combinations for this placement and the rest 4 can be arranged in 4! ways
hence
360
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Re: Arrangements(Combinatorics): Six mobsters for Goodbuddies [#permalink] New post 03 Oct 2011, 17:32
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

Last edited by Bull78 on 05 Oct 2011, 21:24, edited 1 time in total.
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Re: PS: Combinatorics [#permalink] New post 03 Oct 2011, 18:04
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] New post 12 Oct 2011, 13:12
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, 13:12
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