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Re: Six mobsters have arrived at the theater
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10 Aug 2015, 07:39
VenoMfTw wrote: Six mobsters have arrived at the theater for the premiere of a film. One of the mobsters, Eden, is an informer, and he is afraid that another member of his crew, Costa, is on to him. Eden wanting to keep Costa in his sights, insists upon standing behind Costa in line at the concession stand. how many ways can the six arrange themselves in line such that Eden's requirement is satisfied?
a. 6 b. 24 c. 120 d. 360 e. 720 Half of the total possible ways Costa will be ahead of Eden And for other half Eden will be ahead of Costa Total possible ways = 6! = 720 ways Thus for required = 720/2 =360
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Re: Six mobsters have arrived at the theater for the premiere of
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31 Aug 2015, 20:38
Now the method I'm discussing has already been shared, but the explanations were not too clear for me, so sharing a detailed explanation. Now following can be the available arrangements (please note that the question requires us to place Frankie behind Joey in line, though not necessarily right behind him): _ _ _ _ _ F: when F is at the last, there are 5 options for J and the rest can be arranged in 4! ways: 5x4! _ _ _ _ F _: when F is at the second last position, there are 4 options for J and the rest can be arranged in 4! ways: 4x4! _ _ _ F _ _: when F is at the third last position, there are 3 options for J and the rest can be arranged in 4! ways: 3x4! _ _ F _ _ _: when F is at the third third spot from the front, there are 2 options for J and the rest can be arranged in 4! ways: 2x4! _ F _ _ _: when F is at the third third spot from the front, there is only 1 options for J and the rest can be arranged in 4! ways: 4! Now F can not be placed at the first spot as he has to be behind J, so above are al the situations possible and the sum would give us the total arrangements. total arrangements = 5x4! + 4x4! + 3x4! + 2x4! + 4! = (5+4+3+2+1)4! = 15x4! = 360



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Six mobsters have arrived at the theater for the premiere of
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01 Sep 2015, 02:27
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? A. 6 B. 24 C. 120 D. 360 E. 720
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Re: Six mobsters have arrived at the theater for the premiere of
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24 May 2016, 13:51
reply2spg 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?
A. 6 B. 24 C. 120 D. 360 E. 720 We can pick 2 place for Frankie and Joey (2 of 6) and multiply by the other 4! possiblity for the rest. (6!/2!*4!)*4!=360.



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Re: Six mobsters have arrived at the theater for the premiere of
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07 Jul 2017, 11:29
If they could seat anywhere the answer would be 6! = 720 If one had to seat right behind the other it would be 5! = 120
But we know that there are variations on which other people are seating between them, so the answer must be 120 < x < 720
The only option that fits is D (360)



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Re: Six mobsters have arrived at the theater for the premiere of
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01 Jan 2018, 02:56
Hi Bunuel. You have got the best solution for this. However, I approached this problem differently. My method:
Frankie & Joey choose 2 spots in 6C2 ways. And no 2! because they can only arrange in 1 way, F behind J. Remaining spots are taken by 4 in 4! ways.
Therefore, 6C2*4! = 6!/2! = 360



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Re: Six mobsters have arrived at the theater for the premiere of
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05 Feb 2018, 20:33
LevFin7S wrote: for a more complicated way
assume following scenarios where joey stands 1st, 2nd, 3rd, 4, 5 in line
then Frankie can stand on (5 + 4 + 3 + 2 + 1) spots
multiply that by 4! for the other 4 people
15 x 4! = 360 =========================== Thanks for explaining this in a simple way. I realized my mistake. what i did i assumed that let joey stands 1st, 2nd, 3rd, 4, 5 in line then calculated [5!+4!+3!+2!+1!=120]. now i realize that this way I am arranging only people standing behind Joe.



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Re: Six mobsters have arrived at the theater for the premiere of
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12 Aug 2018, 21:04
BunuelMy take  Total 6 places to fill. _ _ _ _ _ J. ( No. Of place for F is 5 OR F can take any place among the 5) _ _ _ _ J _ .( Places for F = 4) Similarly F will have 3,2,1 places as per J position. Total ways = 5*4*3*2*1=120. Where did I go wrong? Please explain ! chetan2uPosted from my mobile device



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Re: Six mobsters have arrived at the theater for the premiere of
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12 Aug 2018, 22:18
Bunuel wrote: reply2spg 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 Please explain Total arrangement of 6 = 6!. In half of the cases Frankie will be behind Joey and in half of the cases Joey will be behind Frankie (as probability doesn't favor any of them). So, the needed arrangement is 6!/2=360. Answer: D (360) Hi Bunuel, Will this change if question it was stated Frankie was supposed to stand right after Joey , will we be using 6C2 in that case ?
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Re: Six mobsters have arrived at the theater for the premiere of
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13 Aug 2018, 09:06
AdityaHongunti wrote: BunuelMy take  Total 6 places to fill. _ _ _ _ _ J. ( No. Of place for F is 5 OR F can take any place among the 5) _ _ _ _ J _ .( Places for F = 4) Similarly F will have 3,2,1 places as per J position. Total ways = 5*4*3*2*1=120. Where did I go wrong? Please explain ! chetan2uPosted from my mobile device You are not taking the combinations in which others can stand... Say I wanted to do as per your way.. 1) _,_,_,_,_, J So remaining 5 in 5! 2) _,_,_,_, J,_ F can take any of 4, and the remaining 4 seats can be taken in 4! Ways So 4*4! 3) _,_,_, J,_,_ Similarly 3*4! 4) _,_, J,_,_,_ So 2*4! 5) _, J,_,_,_,_ So 1*4! Total 5!+4*4!+3*4!+2*4!+1*4!=4!(5+4+3+2+1)=24*15=360.. Hope it helps
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Re: Six mobsters have arrived at the theater for the premiere of &nbs
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