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# Six mobsters have arrived at the theater for the premiere of

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Re: PS: Combinatorics [#permalink]

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18 Mar 2008, 21:43
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, $$N=\frac{720}{2}=360$$
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Re: PS: Combinatorics [#permalink]

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18 Mar 2008, 22:09
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???
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Re: PS: Combinatorics [#permalink]

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18 Mar 2008, 22:17
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
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Re: PS: Combinatorics [#permalink]

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19 Mar 2008, 06:45
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
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Re: Combinations Mobsters - MGMAT [#permalink]

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18 Aug 2008, 17:43
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 1-5 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
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Re: Combinations Mobsters - MGMAT [#permalink]

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21 Aug 2008, 07:21
There is an easier way to do this.
With 6 people, the total permutations will be 6!= 720.
In exactly 1/2 of these instances, Frankie will be in front of Joey and in the other half, Joey will be in front of Frankie.
Therefore, the answer is 720/2 = 360.
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Re: Combinations Mobsters - MGMAT [#permalink]

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21 Aug 2008, 07:41
KASSALMD wrote:
There is an easier way to do this.
With 6 people, the total permutations will be 6!= 720.
In exactly 1/2 of these instances, Frankie will be in front of Joey and in the other half, Joey will be in front of Frankie.
Therefore, the answer is 720/2 = 360.

That is the OE from MGMAT that the poster didn't understand.
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Re: Combinations Mobsters - MGMAT [#permalink]

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21 Aug 2008, 08:24
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 1395 times ]

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Re: Mobsters going for a movie [#permalink]

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13 Sep 2008, 18:14
2
KUDOS
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 (6-joey-frankie); 4! for rest
3. Joey at 3; 4*3 choices for 1-2; 3! for rest
4. Joey at 4; 4*3*2 choices for 1-3; 2! for rest
5. Joey at 5; 4*3*2*1 choice for 1-4; 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.
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Re: Mobsters going for a movie [#permalink]

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13 Sep 2008, 18:25
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
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Re: arrangment problem [#permalink]

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18 Jul 2009, 11:13
1
KUDOS
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
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09 Sep 2009, 15:40
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)
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27 Sep 2009, 21:42
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
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Re: 700-800 Combinatorics hard concept [#permalink]

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05 Jul 2010, 12:27
2
KUDOS

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.
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Re: 700-800 Combinatorics hard concept [#permalink]

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05 Jul 2010, 14:15
1
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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!
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Re: 700-800 Combinatorics hard concept [#permalink]

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05 Jul 2010, 22:35
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 F-6, 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 F-6 => 5 * 4!...Similarly
F-5 => 4 * 4!
..........
F-2 => 1 * 4!

F cannot be positioned as first, hence F-1 is not valid!

Add all of the above
4! * [ 5 + 4 + 3 + 2 + 1 ] = 4! * 15 = 25 * 15 = 360.
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Re: Permutation & Combination [#permalink]

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03 Apr 2011, 16:26
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.
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Re: Permutation & Combination [#permalink]

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03 Apr 2011, 16:52
1
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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

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Re: Permutation & Combination [#permalink]

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23 Apr 2011, 09:27
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.
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Re: Permutation & Combination [#permalink]

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23 Apr 2011, 12:34
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:
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.
Re: Permutation & Combination   [#permalink] 23 Apr 2011, 12:34

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