Author 
Message 
CEO
Joined: 17 Nov 2007
Posts: 3458
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth)  Class of 2011

Re: PS: Combinatorics
[#permalink]
Show Tags
18 Mar 2008, 22:43
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, \(N=\frac{720}{2}=360\)
_________________
HOT! GMAT TOOLKIT 2 (iOS) / GMAT TOOLKIT (Android)  The OFFICIAL GMAT CLUB PREP APP, a musthave app especially if you aim at 700+  Limited GMAT/GRE Math tutoring in Chicago



Manager
Joined: 02 Jan 2008
Posts: 190
Location: Toronto

Re: PS: Combinatorics
[#permalink]
Show Tags
18 Mar 2008, 23:09
Yes, you guys are right.
I understood the problem wrong. I thought if Frankie wanted to only stand behind joey then
if Joey is first place in the line, Frankie could have any 2,3,4,5 or 6 th spot. (5!) If Joey is 2nd place in the row, Frankie could have 3,4,5 or 6th (4!) If Joey is 3rd place in the row, Frankie could have 4,5 or 6 (3!) .............4th..................................................5 or 6 (2!) .............5th...................................................6 (1 option)
What is wrong with this logics???



CEO
Joined: 17 Nov 2007
Posts: 3458
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth)  Class of 2011

Re: PS: Combinatorics
[#permalink]
Show Tags
18 Mar 2008, 23:17
maverick101 wrote: Yes, you guys are right.
I understood the problem wrong. I thought if Frankie wanted to only stand behind joey then
if Joey is first place in the line, Frankie could have any 2,3,4,5 or 6 th spot. (5!=5*4!) If Joey is 2nd place in the row, Frankie could have 3,4,5 or 6th (4!  4*4!) If Joey is 3rd place in the row, Frankie could have 4,5 or 6 (3!  3*4!) .............4th..................................................5 or 6 (2!  2*4!) .............5th...................................................6 (1 option  1*4!)
What is wrong with this logics??? You forgot that we always have 4! possibilities for others when we fixed positions of Joey and Frankie
_________________
HOT! GMAT TOOLKIT 2 (iOS) / GMAT TOOLKIT (Android)  The OFFICIAL GMAT CLUB PREP APP, a musthave app especially if you aim at 700+  Limited GMAT/GRE Math tutoring in Chicago



SVP
Joined: 29 Mar 2007
Posts: 2465

Re: PS: Combinatorics
[#permalink]
Show Tags
19 Mar 2008, 07:45
maverick101 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?
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. I first tried doin the long approach: FJXXXX, FXJXXXX etc... but seems this approach is much faster. 6!=720 total number of ways with no restraints. F will be behind J 1/2 of the times. Thus answer should be 360



Manager
Joined: 15 Jul 2008
Posts: 205

Re: Combinations Mobsters  MGMAT
[#permalink]
Show Tags
18 Aug 2008, 18:43
jlola21 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 MGMAT's explanation I didn't quite understand, it was very brief. Can someone please explain this? F always has to stand behind J. Given that, J can stand at any of the 15 positions. When.... J is at #1  The remaining 5 can be arranged in 5! = 120 ways J is at #2  #1 can be filled by any of the 4 except F. The remaining 4 behind J can be arranged in 4! ways. So total arrangements in this case 4*4! = 96 ways J is at #3  first and second positions can be filled by 2 out of 4 (except F) in 4P2 ways. The remaining 3 behind J can be arranged in 3! ways. So total arrangements in this case 4P2*3! = 72 ways J is at #4  first second and third positions can be filled by 3 out of 4 (except F) in 4P3 ways. The remaining 2 behind J can be arranged in 2! ways. So total arrangements in this case 4P3*2! = 48 ways J is at #5  The first 4 position can be filled in 4P4 ways. Only F stands behind J. So total number of arrangements for this one is 4P4 =4!=24 ways Total number of ways = 120+96+72+48+24 = 360 ways



Manager
Joined: 30 Jul 2007
Posts: 130

Re: Combinations Mobsters  MGMAT
[#permalink]
Show Tags
21 Aug 2008, 09:24
Visually (see attachment): 15 F and J combinations, and for the remaining 4 mobsters: 4! = 24 so in total: 15 X 24 = 360
Attachments
FJ.JPG [ 12.59 KiB  Viewed 1790 times ]



Manager
Joined: 10 Apr 2008
Posts: 53

Re: Mobsters going for a movie
[#permalink]
Show Tags
13 Sep 2008, 19:14
Here's how I did it, but it seems long and cumbersome so hopefully someone can provide a better explanation.
There are 6 possible positions for each mobster. Frankie depends on Joey so select those first
1. Joey at 1; 5! for rest 2. Joey at 2; 4 choices for 1 (6joeyfrankie); 4! for rest 3. Joey at 3; 4*3 choices for 12; 3! for rest 4. Joey at 4; 4*3*2 choices for 13; 2! for rest 5. Joey at 5; 4*3*2*1 choice for 14; 1 for rest
So total = 5! + 4x4! + 4*3*3! + 4*3*2*2! + 4! = 5(4!) + 4(4!) + 3(4!) + 2(4!) + (4!) = 4! x (5 + 4 + 3 + 2 + 1) = 24 * 15 = 360
So D.



Intern
Joined: 26 Aug 2008
Posts: 15

Re: Mobsters going for a movie
[#permalink]
Show Tags
13 Sep 2008, 19:25
6 guys can seat in 6! different ways. 6x5x4x3x2x1 = 720 ways what are the chances that these two guys will sit one after another ? 50 % 720/2 = 360 D krishan 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



Senior Manager
Joined: 04 Jun 2008
Posts: 270

Re: arrangment problem
[#permalink]
Show Tags
18 Jul 2009, 12:13
F must be behind J and the others 4 can stand in any way.
Say J stands in 1st place, ie, in front of all others. Then F can stand anywhere behind in 5 ways, and others in !4 ways.
If J stands in 2nd place, F can stand in any of the remaining 4 places behind J in 4 ways, and all others can again arrange in !4
J  3rd place, F  stand in 3 places in 3 ways and others in !4....
J in 4th and 5th, F in 2 and 1 way respectively, and others in !4 each time.
So total number of ways is by adding all of the above.
!4 (5+4+3+2+1) = 360



Intern
Joined: 07 Aug 2009
Posts: 46

Re: Combination
[#permalink]
Show Tags
09 Sep 2009, 16:40
bmwhype2 wrote: 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!? nah it wldnt be 5!2!, it would just be 5! which is 120. this is because frankie must always stand behind the other guy( joey?? dnt rembr the name)



Manager
Joined: 27 Oct 2008
Posts: 180

Re: Combination
[#permalink]
Show Tags
27 Sep 2009, 22:42
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
Soln: Ans is D
To find the total number of arrangements in which Frankie comes behind Joey, we need to fix Frankie to each of the places and see the possibilities of arranging Joey. = 5! + 4 * 4! + 12 * 3! + 24 * 2! + 4! = 360



Manager
Joined: 04 Feb 2010
Posts: 165

Re: 700800 Combinatorics hard concept
[#permalink]
Show Tags
05 Jul 2010, 13:27
Here's the answer.
There are fifteen ways F would be behind J all the time. Think of it this way. There are six places, and you have to select two places for F & J, and where the other four's positions does not matter. This can be put into numbers as 6!/4!2! giving us 15.
So you'd have the following:
FJ????, F?J???, F??J??, F???J?, F????J ?FJ???, ?F?J??, ?F??J?, ?F???J ??FJ??, ??F?J?, ??F??J ???FJ?, ???F?J ????FJ
15 scenarios in all.
Now, you can use counting techniques to get all the possible orders of the four remaining mobsters. That would be 4! = 24.
Multiply the 15 cases with all of the possible orders of the four remaining mobsters  15*24 = 360.



Kaplan GMAT Instructor
Joined: 21 Jun 2010
Posts: 71
Location: Toronto

Re: 700800 Combinatorics hard concept
[#permalink]
Show Tags
05 Jul 2010, 15:15
Quote: I have a hard time grasping the concept that half of the time F would be behind J. Can someone please explain this to me? Hi, maybe this will help: take 2 objects, x and y. How many ways to arrange them? Clearly, 2: xy and yx. Notice that in half the arrangements (1 out of 2) x is before y, and in the other half y is before x. Take 3 objects, x, w, and y. There are 3! or 6 arrangements. In half (ie, 3) of those arrangements, x is before w, while in the other half w before x. Likewise, in half of the arrangements x is before y while in the other half y before x. And, also, in half the arrangements, w is before y while in the other half y before w. Why would it be the case that Joey can be arranged ahead of Frankie more or less often than Frankie can be arranged ahead of Joey? Why not the other way around? So, here, the easiest way to solve is certainly to take just half of the total arrangements: 6!/2 = 360, and choose D. Quote: Also, can this be down using straight combinations without getting the concept that half of the time F would be behind J? Yes, it certainly can as the above poster demonstrated! But many combinatorics questions on the GMAT resist pure formulaic treatment. A little bit of reasoning on these questions can save you an immense amount of time!



Intern
Joined: 03 Jun 2010
Posts: 22

Re: 700800 Combinatorics hard concept
[#permalink]
Show Tags
05 Jul 2010, 23:35
knabi wrote: I have a hard time grasping the concept that half of the time F would be behind J. Can someone please explain this to me? Also, can this be down using straight combinations without getting the concept that half of the time F would be behind J? Thanks.
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 J is always before F If F in placed in 6th position F6, then J can be placed in one of the five positions (1,2,3,4,5) hence five ways. Other 4 persons can be positioned in 4! ways, hence F6 => 5 * 4!...Similarly F5 => 4 * 4! .......... F2 => 1 * 4! F cannot be positioned as first, hence F1 is not valid! Add all of the above 4! * [ 5 + 4 + 3 + 2 + 1 ] = 4! * 15 = 25 * 15 = 360.



Intern
Joined: 23 Feb 2011
Posts: 3

Re: Permutation & Combination
[#permalink]
Show Tags
03 Apr 2011, 17:26
Here's how I solved it; hope this helps:
I added up all the different combinations for Frankie to be behind Joey.
1st scenario: Joey is first, so there is one option for the first spot, and 5! options for the remaining spot: 1 * 5 * 4 * 3 * 2 * 1 = 120
2nd scenario: Joey is second, so there is one option for the 2nd spot, 4 options for the first spot (1st spot can't be joey or frankie, so must be on of the other 4) and 4! options for the remaining spots (out of 6 possible people, we've used up Joey already and 1 of the other guys for the first seat, so there are 4 guys remaining to arrange: 4 * 1 * 4 * 3 *2 * 1 = 96
3rd scenario: Joey is 3rd, so there is one option for the third spot, 4 options for the 1st spot (can't be Joey or Frankie), 3 options for the second spot (can't be J, F, or the guy in spot 1), and 3! for the remaining spots (can't be J or the two guys up front): 4*3*1*3*2*1 = 72
4th scenario: J is in 4th spot, so there is one option for that spot, first three spots are 4*3*2 (based on logic above), and there are two options for the 5th spot (can't be J or the three guys up front) 4*3*2*1*2*1= 48
5th scenario: J is 5th, so there is one option for that spot, F only has one place he can go, 6th, so there is one option for that spot, the other guys are permutated up front: 4*3*2*1*1*1 =24
6th scenario: there is no 6th scenario as J can't be last!
Add up all the permutations: 120+96+72+48+24 = 360
The explanation looks long, but if it clicks, you can solve the problem using this method in < 2 minutes.



Retired Moderator
Joined: 16 Nov 2010
Posts: 1451
Location: United States (IN)
Concentration: Strategy, Technology

Re: Permutation & Combination
[#permalink]
Show Tags
03 Apr 2011, 17:52
J in 1 then the remaining in 5! ways = 120 J in 2 then F in 4 ways and the remaning in 4! ways = 24 * 4 = 96 J in 3 then F in 3 ways and remaining in 4! way = 24 * 3 = 72 J in 4 then F in 2 ways and remaining in 4! way = 24 * 2 = 48 J in 5 then F in 1 way and remaining in 4! way = 24 * 1 = 24 So total = 120 + 96 + 72 + 48 + 24 = 360 Answer  D
_________________
Formula of Life > Achievement/Potential = k * Happiness (where k is a constant)
GMAT Club Premium Membership  big benefits and savings



Director
Joined: 01 Feb 2011
Posts: 670

Re: Permutation & Combination
[#permalink]
Show Tags
23 Apr 2011, 13:34
usually without restrictions you can arrange these people in 6! ways . in half of the situations F would be behind J and in other way F would be ahead of J. here we are only interested in arrangements where F would be behind J. Thats the reason we are dividing by 2 ( to get rid of the other half of the arrangements where F is behind J). Hence answer is 6!/2 = 360 hope its clears now. dumluck wrote: reply2spg wrote: insists upon standing behind Joey in line at the concession stand, though not necessarily right behind him. 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.



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 8293
Location: Pune, India

Re: Permutation & Combination
[#permalink]
Show Tags
24 Apr 2011, 16:09
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.
_________________
Karishma Veritas Prep GMAT Instructor
Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >
GMAT selfstudy has never been more personalized or more fun. Try ORION Free!



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 8293
Location: Pune, India

Re: PS  700 level  arrangements
[#permalink]
Show Tags
04 Sep 2011, 21:06
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: advancedconstraintcombinatorics42275.html
_________________
Karishma Veritas Prep GMAT Instructor
Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >
GMAT selfstudy has never been more personalized or more fun. Try ORION Free!



Manager
Status: Do till 740 :)
Joined: 13 Jun 2011
Posts: 92
Concentration: Strategy, General Management
GPA: 3.6
WE: Consulting (Computer Software)

Re: Arrangements(Combinatorics): Six mobsters for Goodbuddies
[#permalink]
Show Tags
14 Sep 2011, 12:38
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




Re: Arrangements(Combinatorics): Six mobsters for Goodbuddies &nbs
[#permalink]
14 Sep 2011, 12:38



Go to page
Previous
1 2 3
Next
[ 44 posts ]



