## 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...
"When you want to succeed as bad as you want to breathe, then you'll be successful." - Eric Thomas

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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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