Author 
Message 
TAGS:

Hide Tags

Manager
Joined: 14 Mar 2014
Posts: 147

Re: Six mobsters have arrived at the theater
[#permalink]
Show Tags
10 Aug 2015, 08:39
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
_________________
I'm happy, if I make math for you slightly clearer And yes, I like kudos ¯\_(ツ)_/¯



Intern
Joined: 11 Dec 2012
Posts: 31

Six mobsters have arrived at the theater for the premiere of
[#permalink]
Show Tags
01 Sep 2015, 03:27
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
Attachments
Six Mobsters.JPG [ 59.89 KiB  Viewed 1179 times ]



Current Student
Joined: 18 Sep 2015
Posts: 93
GMAT 1: 610 Q43 V31 GMAT 2: 610 Q47 V27 GMAT 3: 650 Q48 V31 GMAT 4: 700 Q49 V35
WE: Project Management (Health Care)

Re: Six mobsters have arrived at the theater for the premiere of
[#permalink]
Show Tags
24 May 2016, 14:51
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.



Intern
Joined: 01 Sep 2016
Posts: 8

Re: Six mobsters have arrived at the theater for the premiere of
[#permalink]
Show Tags
07 Jul 2017, 12:29
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)



Manager
Joined: 10 Dec 2011
Posts: 102
Location: India
Concentration: Finance, Economics
GMAT Date: 09282012
WE: Accounting (Manufacturing)

Re: Six mobsters have arrived at the theater for the premiere of
[#permalink]
Show Tags
01 Jan 2018, 03:56
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



Intern
Joined: 03 Jul 2015
Posts: 9

Re: Six mobsters have arrived at the theater for the premiere of
[#permalink]
Show Tags
05 Feb 2018, 21:33
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.



Manager
Joined: 20 Sep 2016
Posts: 191

Re: Six mobsters have arrived at the theater for the premiere of
[#permalink]
Show Tags
12 Aug 2018, 22:04
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



Manager
Joined: 26 Feb 2018
Posts: 71
WE: Sales (Internet and New Media)

Re: Six mobsters have arrived at the theater for the premiere of
[#permalink]
Show Tags
12 Aug 2018, 23:18
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 ?
_________________
" Can't stop learning and failing"



Math Expert
Joined: 02 Aug 2009
Posts: 6554

Re: Six mobsters have arrived at the theater for the premiere of
[#permalink]
Show Tags
13 Aug 2018, 10:06
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
_________________
1) Absolute modulus : http://gmatclub.com/forum/absolutemodulusabetterunderstanding210849.html#p1622372 2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html 3) effects of arithmetic operations : https://gmatclub.com/forum/effectsofarithmeticoperationsonfractions269413.html
GMAT online Tutor




Re: Six mobsters have arrived at the theater for the premiere of &nbs
[#permalink]
13 Aug 2018, 10:06



Go to page
Previous
1 2
[ 30 posts ]



