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Mobster combinatorics

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Mobster combinatorics

by szDave » Mon Jan 28, 2013 7:36 am
Hello all,

Found a good one!

Six mobsters have arrived atthe theater forthe premiere of the film Goodbuddies. One ofthe mobsters, Frankie, is an informer, and hes 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 Frankies requirement is satisfied?

It's funny I calculated the answer, but took 3:51!
Can you give me some explanation, thanks!
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by Brent@GMATPrepNow » Mon Jan 28, 2013 7:41 am
szDave wrote:Hello all,

Found a good one!

Six mobsters have arrived atthe theater forthe premiere of the film Goodbuddies. One ofthe mobsters, Frankie, is an informer, and hes 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 Frankies requirement is satisfied?
If we remove the restriction that Frankie must stand behind Joey, then the mobsters can be arranged in 6! (720) different ways.

Now, of these 720 different arrangements, it is clear that Frankie will be standing behind Joey half the time, and Joey will be standing behind Frankie half the time. (in other words, for every arrangement where Frankie is ahead of Joey, there is an arrangement where Joey is ahead of Frankie).

So, the answer = 720/2 = 360.

Extension: If there were n mobsters altogether, then the answer would be n!/2

Cheers,
Brent
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by vomhorizon » Mon Jan 28, 2013 7:55 am
This is an MGMAT question that i also spent a lot of time when i wrote the CAT. My problem was that i did not give the usual 10-12 seconds to analyze the question (which i believe is a must for Probability and P & C question) and hit the calculations straight after reading..

The logic that you have to grasp here is that the mobsters are going to standing in a line and no matter the arrangement no 2 will be standing next to each other. So logically half the times one would be behind another mobster and the other half ahead ... Given this logic all you have to do is calculate the total no of ways 6 mobsters can stand in a line ie. 6! and halve that , so 6!/2 , 720/2 = 360...
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by szDave » Mon Jan 28, 2013 8:31 am
Thanks for the shortener. I calculated it like this:

slot method, 2 stand behind each other: then there is 4*3*2*1 ways for this, then a simple combination solves how many ways can the 2 in the 6 spaces arranged, that is : 6!/(2!*4!) - 24*15=360.
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by Brent@GMATPrepNow » Mon Jan 28, 2013 8:44 am
szDave wrote:Thanks for the shortener. I calculated it like this:

slot method, 2 stand behind each other: then there is 4*3*2*1 ways for this, then a simple combination solves how many ways can the 2 in the 6 spaces arranged, that is : 6!/(2!*4!) - 24*15=360.
Looks good!

For others out there, I thought I'd show the "slot" method step-by-step:

Take the task of arranging all 6 mobsters and break it into stages.

Stage 1: Select 2 spaces for Frankie and Joey

NOTE: once we've selected the 2 spaces, Frankie must stand behind Joey. So, once we've selected the 2 spaces, there's only one suitable arrangement for Frankie and Joey.

Since the order of the 2 selected spaces does not matter (for example, selecting space#1 then space#3 is the same as selecting space#3 then space#1), we can use combinations.
There are 6 spaces and we must select 2.
This can be accomplished in 6C2 ways (= 15 ways).

Aside: If anyone is interested, we have a free video on calculating combinations (like 6C2) in your head: https://www.gmatprepnow.com/module/gmat-counting?id=789

Stage 2: Arrange one of the remaining 4 mobsters.
There are 4 spaces remaining, so we can complete this stage in 4 ways.

Stage 3: Arrange one of the remaining 3 mobsters.
There are 3 spaces remaining, so we can complete this stage in 3 ways.

Stage 4: Arrange one of the remaining 2 mobsters.
There are 2 spaces remaining, so we can complete this stage in 2 ways.

Stage 5: Arrange the last remaining mobster.
There is 1 space remaining, so we can complete this stage in 1 way.

By the Fundamental Counting Principle (FCP) we can complete all 5 stages (and thus arrange all 6 mobsters) in (15)(4)(3)(2)(1) ways ([spoiler]= 360 ways[/spoiler])

Cheers,
Brent

Aside: For more information about the FCP, we have a free video on the subject: https://www.gmatprepnow.com/module/gmat-counting?id=775
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by szDave » Mon Jan 28, 2013 8:57 am
Great explanation!
Yes, this problem was in MGMAT cat test.
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by DBushkalov » Wed Feb 27, 2013 11:30 am
Brent@GMATPrepNow wrote:
szDave wrote:
Stage 1:


Since the order of the 2 selected spaces does not matter (for example, selecting space#1 then space#3 is the same as selecting space#3 then space#1), we can use combinations.
There are 6 spaces and we must select 2.
This can be accomplished in 6C2 ways (= 15 ways).

https://www.gmatprepnow.com/module/gmat-counting?id=775

hi Brent,

let A B C D E F represent the 6 mobster. The only 2 that matter are F and E: (F= frankie, E= the one frankie is afraid of)


i want to know if I got this right. I thing you use combinations insteadof permutations, in order to only have the combination A B C D F E (frankfie is behind E) AND TO exclude the Permutation A B C D E F (the one in which E is behind F, or frankie is ahed of the bad mobster- E).

is that it ?
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by Brent@GMATPrepNow » Wed Feb 27, 2013 11:44 am
DBushkalov wrote:
Brent@GMATPrepNow wrote:
szDave wrote:
Stage 1:


Since the order of the 2 selected spaces does not matter (for example, selecting space#1 then space#3 is the same as selecting space#3 then space#1), we can use combinations.
There are 6 spaces and we must select 2.
This can be accomplished in 6C2 ways (= 15 ways).

https://www.gmatprepnow.com/module/gmat-counting?id=775

hi Brent,

let A B C D E F represent the 6 mobster. The only 2 that matter are F and E: (F= frankie, E= the one frankie is afraid of)


i want to know if I got this right. I thing you use combinations insteadof permutations, in order to only have the combination A B C D F E (frankfie is behind E) AND TO exclude the Permutation A B C D E F (the one in which E is behind F, or frankie is ahed of the bad mobster- E).

is that it ?
When people first see this question, it seems that order is very important here. After all, Frankie insists upon standing behind Joey in line. However, in our solutions, depending on how we set them up, order may or may not matter.

In my first solution, order did matter since I was arranging all 6 mobsters.

The second solution is a little different. In stage 1, I'm selecting two spaces, that's it (I'm not selecting Mobsters and I'm not arranging Mobsters). Once I select those two spaces, then I'll deal with placing Frankie and Joey in those spaces (at which time, we see that there's only 1 way to place them).
So, if I'm just selecting 2 spaces, the order does not matter.


I hope that helps.

Cheers,
Brent
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by gmattesttaker2 » Sun Mar 03, 2013 11:06 pm
Brent@GMATPrepNow wrote:
DBushkalov wrote:
Brent@GMATPrepNow wrote:
szDave wrote:
Stage 1:


Since the order of the 2 selected spaces does not matter (for example, selecting space#1 then space#3 is the same as selecting space#3 then space#1), we can use combinations.
There are 6 spaces and we must select 2.
This can be accomplished in 6C2 ways (= 15 ways).

https://www.gmatprepnow.com/module/gmat-counting?id=775

hi Brent,

let A B C D E F represent the 6 mobster. The only 2 that matter are F and E: (F= frankie, E= the one frankie is afraid of)


i want to know if I got this right. I thing you use combinations insteadof permutations, in order to only have the combination A B C D F E (frankfie is behind E) AND TO exclude the Permutation A B C D E F (the one in which E is behind F, or frankie is ahed of the bad mobster- E).

is that it ?
When people first see this question, it seems that order is very important here. After all, Frankie insists upon standing behind Joey in line. However, in our solutions, depending on how we set them up, order may or may not matter.

In my first solution, order did matter since I was arranging all 6 mobsters.

The second solution is a little different. In stage 1, I'm selecting two spaces, that's it (I'm not selecting Mobsters and I'm not arranging Mobsters). Once I select those two spaces, then I'll deal with placing Frankie and Joey in those spaces (at which time, we see that there's only 1 way to place them).
So, if I'm just selecting 2 spaces, the order does not matter.


I hope that helps.

Cheers,
Brent

Hello Brent,

Can you please help me? I was trying to use the following approach:

One possible arrangement is Joey Frank Other Other Other Other

i.e. P(JFOOOO) = (1/6)(1/5)(4/4)(3/3)(2/2)(1/1) = 1/30

Now I try to find out the total number of ways I can arrange these 6 objects with the criteria that Frankie stands behind Joey

(5 * 5 * 4 * 3 * 2 * 1 )/4!

= 25

I was just wondering if this approach is correct here. I was trying to solve more on the approach of this problem:

https://www.beatthegmat.com/tricky-proba ... tml#485339

Thank you very much for all your valuable time and help.

Best Regards,
Sri
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by Brent@GMATPrepNow » Mon Mar 04, 2013 8:23 am
gmattesttaker2 wrote: Hello Brent,

Can you please help me? I was trying to use the following approach:

One possible arrangement is Joey Frank Other Other Other Other

i.e. P(JFOOOO) = (1/6)(1/5)(4/4)(3/3)(2/2)(1/1) = 1/30

Now I try to find out the total number of ways I can arrange these 6 objects with the criteria that Frankie stands behind Joey

(5 * 5 * 4 * 3 * 2 * 1 )/4!

= 25

I was just wondering if this approach is correct here. I was trying to solve more on the approach of this problem:

https://www.beatthegmat.com/tricky-proba ... tml#485339

Thank you very much for all your valuable time and help.

Best Regards,
Sri
Hi Sri,

There are a few puzzling things about your solution.
First there's this part: P(JFOOOO) = (1/6)(1/5)(4/4)(3/3)(2/2)(1/1) = 1/30

I'm not sure why you've injected some probability into this, when the question is strictly a counting question.

Now, your strategy of counting the arrangements of Joey, Frank, Other, Other, Other and Other will work only if the question had stated something to the effect that the other 4 people are all identical.
If that were the case, then you could apply the MISSISSIPPI rule here.
However, there's no indication that the question suggests that the other 4 people are identical.
For example, the arrangement Joey, Frankie, Ken, Al, Bob, Don is different from the arrangement Joey, Frankie, Ken, Bob, Al,Don
In your solution, the two arrangements are considered equal.

I hope that helps.

Cheers,
Brent
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