Author 
Message 
TAGS:

Hide Tags

Director
Joined: 16 May 2007
Posts: 548

Six mobsters have arrived at the theater for the premiere of [#permalink]
Show Tags
24 Jul 2007, 07:08
11
This post was BOOKMARKED
Question Stats:
50% (01:44) correct
50% (00:30) wrong based on 242 sessions
HideShow timer Statistics
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. 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 OPEN DISCUSSION OF THIS QUESTION IS HERE: sixmobstershavearrivedatthetheaterforthepremiereofthe126151.html
Official Answer and Stats are available only to registered users. Register/ Login.



Senior Manager
Joined: 04 Jun 2007
Posts: 345

Re: PS: Mobsters [#permalink]
Show Tags
24 Jul 2007, 07:13
trahul4 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. 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
Since F and J are to be always together, number of arrangements is 5! (considering JF as one). Hence, 120 is the answer.
Here is a discussion thread for the same problem:
http://www.gmatclub.com/phpbb/viewtopic.php?t=45097&highlight=mobster



GMAT Club Legend
Joined: 07 Jul 2004
Posts: 5043
Location: Singapore

Since Frankie and Joey must always be in a certain order, we can take them as one entity.
So the number of ways to arrange them = 5! = 120



Director
Joined: 16 May 2007
Posts: 548

2
This post was BOOKMARKED
Yes looks very easy problem.
But OA is 360. I got this question in one of the MGMAT CAT and was surprised to see 120 worng. Anyways here is the explaination given by MGMAT.
Ignoring Frankie's requirement for a moment, observe that the six mobsters can be arranged 6! or 6 x 5 x 4 x 3 x 2 x 1 = 720 different ways in the concession stand line. In each of those 720 arrangements, Frankie must be either ahead of or behind Joey. Logically, since the combinations favor neither Frankie nor Joey, each would be behind the other in precisely half of the arrangements. Therefore, in order to satisfy Frankie's requirement, the six mobsters could be arranged in 720/2 = 360 different ways.
The correct answer is D.
I feel this soultion given by MGMAT is incorrect.



GMAT Club Legend
Joined: 07 Jul 2004
Posts: 5043
Location: Singapore

trahul4 wrote: Yes looks very easy problem. But OA is 360. I got this question in one of the MGMAT CAT and was surprised to see 120 worng. Anyways here is the explaination given by MGMAT.
Ignoring Frankie's requirement for a moment, observe that the six mobsters can be arranged 6! or 6 x 5 x 4 x 3 x 2 x 1 = 720 different ways in the concession stand line. In each of those 720 arrangements, Frankie must be either ahead of or behind Joey. Logically, since the combinations favor neither Frankie nor Joey, each would be behind the other in precisely half of the arrangements. Therefore, in order to satisfy Frankie's requirement, the six mobsters could be arranged in 720/2 = 360 different ways.
The correct answer is D.
I feel this soultion given by MGMAT is incorrect.
Sounds dodgy.. okay, let's try the longwinded way. We know Frankie always wants to stand behind Joey.
So,
 if Joey is first in the queue, Frankie must be second. The remaining 4 mobsters can be arranged 4! ways.
 if Joey is second in the queue, Frankie must be third. The remaining 4 mobsters can be arranged 4! ways.
 if Joey is third in the queue, Frankie must be fourth. The remaining 4 mobsters can be arranged 4! ways.
 if Joey is fourth in the queue, Frankie must be fifth. The remaining 4 mobsters can be arranged 4! ways.
 if Joey is fifth in the queue, Frankie must be last. The remaining 4 mobsters can be arranged 4! ways.
So total = 5 * 4! = 120
This method should be foolproof as we are just shifting the pair along the queue. Don't see how you can get 360.



Director
Joined: 16 May 2007
Posts: 548

I agree with you. I think the OA given by MGMAT is wrong and my answer on the CAT was correct.



Manager
Joined: 08 Jul 2007
Posts: 173

trahul4 wrote: I agree with you. I think the OA given by MGMAT is wrong and my answer on the CAT was correct.
I had the same problem with MGMAT. This problem has been discussed several times on this board. MGMAT is actually asking how many ways for Frankie to be behind Joey but not directly behind him. Thus, the ways of arranging the 6 people is 6!=720 and Frankie is behind Joey 1/2 of the times so the answer is 360. The answer allows for there to be mobsters between Joey and Frankie.
MGMAT was not clear on this question in my opinion. However, since they did not specify "directly behind" then I guess their answer is okay.



Manager
Joined: 27 May 2007
Posts: 128

I hope the wording isn't this tricky on the actual test! I also got 120 until I read the explanation, assuming behind meant directly behind.



SVP
Joined: 21 Jul 2006
Posts: 1510

Combination [#permalink]
Show Tags
24 Nov 2007, 13:40
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. 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
I really don't know how to approach this problem. Would anyone show me how? appreciate it!



SVP
Joined: 29 Aug 2007
Posts: 2473

Re: Combination [#permalink]
Show Tags
24 Nov 2007, 13:54
1
This post received KUDOS
tarek99 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. 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
I really don't know how to approach this problem. Would anyone show me how? appreciate it!
1. If JF are adjacent to each other = 5! = 120
1. If F is behind but not adjcent to J = 3 x 5! = 360



SVP
Joined: 21 Jul 2006
Posts: 1510

thank you and the OA is 360. here is how i approached it. because J and F must be next to each other, the whole arrangement is 5! then, because they have to be next to each other, i also considered the 2!. so when i multiple 5!*2!, i get 240. yet this approach is wrong. i don't understand why you used just 3. why is it wrong to use the 2! as well? i understand the 5!, but don't understand why you used just 3 instead of the 2!
would you explain? thanks



SVP
Joined: 29 Aug 2007
Posts: 2473

tarek99 wrote: thank you and the OA is 360. here is how i approached it. because J and F must be next to each other, the whole arrangement is 5! then, because they have to be next to each other, i also considered the 2!. so when i multiple 5!*2!, i get 240. yet this approach is wrong. i don't understand why you used just 3. why is it wrong to use the 2! as well? i understand the 5!, but don't understand why you used just 3 instead of the 2!
would you explain? thanks
= 5 x 4! + 4 x 4! + 3 x 4! + 2 x 4! + 4!
= 4! (5+4+3 +2 + 1)
= 4! 15
= 4! x 5x3
= 5! x 3



Senior Manager
Joined: 09 Oct 2007
Posts: 466

D.
(5!)(15/5) = (120)(3)=360
Where 5! represents the ways in which the mobstersJF can sit. 15/5 are the ways in which JF can sit.



SVP
Joined: 29 Aug 2007
Posts: 2473

spider wrote: GMAT TIGER wrote: tarek99 wrote: thank you and the OA is 360. here is how i approached it. because J and F must be next to each other, the whole arrangement is 5! then, because they have to be next to each other, i also considered the 2!. so when i multiple 5!*2!, i get 240. yet this approach is wrong. i don't understand why you used just 3. why is it wrong to use the 2! as well? i understand the 5!, but don't understand why you used just 3 instead of the 2!
would you explain? thanks = 5 x 4! + 4 x 4! + 3 x 4! + 2 x 4! + 4! = 4! (5+4+3 +2 + 1) = 4! 15 = 4! x 5x3 = 5! x 3 I dint get this? anyone please explain!
ok its bit tricky:
1. JF1234 = 5 x 4!
2. 1JF234 = 4 x 4!
3. 12JF34 = 3 x 4!
4. 123JF4 = 2 x 4!
5. 1234JF = 1 x4!
add them up = 5 x 4! + 4 x 4! + 3 x 4! + 2 x 4! + 4! = 5! x 3
hope it is clear now.



CEO
Joined: 17 Nov 2007
Posts: 3584
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth)  Class of 2011

shortcut:
every cases is JF or FJ. Therefore, we should (total number of combination) divide by 2 (due to symmetry)
6!/2=5!*3=360



SVP
Joined: 21 Jul 2006
Posts: 1510

6
This post received KUDOS
I think this is the clearest way to look at it:
JF1234
12JF34
1234JF
Since JF is considered 1 group, the total number of combinations would be 5!. Since there are 3 ways in which JF can be together, we multiply 5! by 3, therefore:
5!*3= 360
hope that's clearer now



Intern
Joined: 01 Dec 2007
Posts: 31

Here is the explanation from MGMT
Ignoring Frankie's requirement for a moment, observe that the six mobsters can be arranged 6! or 6 x 5 x 4 x 3 x 2 x 1 = 720 different ways in the concession stand line. In each of those 720 arrangements, Frankie must be either ahead of or behind Joey. Logically, since the combinations favor neither Frankie nor Joey, each would be behind the other in precisely half of the arrangements. Therefore, in order to satisfy Frankie's requirement, the six mobsters could be arranged in 720/2 = 360 different ways.
The correct answer is D.

Apparently, Frankie and Joey do not need to be right next to each other.



CEO
Joined: 21 Jan 2007
Posts: 2739
Location: New York City

What if it read...
Frankie insists upon standing next to Joey in line at the concession stand. How many ways can the six arrange themselves in line such that Frankie’s requirement is satisfied?
would it be 5! 2!?



Manager
Joined: 02 Jan 2008
Posts: 193
Location: Toronto

PS: Combinatorics [#permalink]
Show Tags
18 Mar 2008, 22:06
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?
I did'nt post the available options on purpose. Please calculate and post your answer and how you got the answer. I will post the available options later. I appologize if this seems incovenient to you.



Senior Manager
Joined: 15 Aug 2007
Posts: 252
Schools: Chicago Booth

Re: PS: Combinatorics [#permalink]
Show Tags
18 Mar 2008, 22:28
11
This post received KUDOS
360
If Joey is in 6th spot, Frank can stand anywhere from 15  rest can be arranged in 4! ways  total 5*4! If Joey is in 5th spot, Frank can stand anywhere from 14  rest can be arranged in 4! ways  total 4*4! . . Total number of arrangments = (1+2+3+4+5)*4! = 360




Re: PS: Combinatorics
[#permalink]
18 Mar 2008, 22:28



Go to page
1 2 3
Next
[ 49 posts ]




