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Counting - Six mobsters

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Counting - Six mobsters

by Brent@GMATPrepNow » Sat Oct 24, 2009 6:46 am
I recently came across this question, and I quite like it.

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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by sanjana » Sat Oct 24, 2009 8:00 am
IMO : 360

If there was no restriction on the arrangement the answer would have been 6! = 720, so our answer is definitely less than 720,eliminate E.

Now,

Imagine the line is as below
Front
-
-
-
-
-
-
Back

Now Joey can be in the 1st position and Frankie could be in any of the other 5 positions to have his rule satisfied.

No of ways : 1 * 5! (Remaining 5 ppl) = 120

Now if Joey is in the 2nd position,
Frankie can take only one of the 4 positions behind him
So Joey occupies his position in 1 way and Frankie has 4 seats to choose from and the remaining 4 people can choose their positions in 4! ways
Hence Total ways : 1*4*4! = 96

Similarly if Joey takes the 3rd
No of ways : 1*3*4! = 72

.
.
.
.
Joey takes the 5th position
Frankie has only 1 choice now
hence : 1*1*4!

Adding all I actually get 350 (120+96+72+48+24). Not sure if I have made a mistake somewhere or whether I am missing something!

Brett,whats the actual answer?
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by dmateer25 » Sat Oct 24, 2009 8:45 am
I agree with Brent; this is a great question!

The answer is indeed 360. However, there is a shortcut to this problem. I won't post it. I will let others give it a shot.
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by NikolayZ » Sat Oct 24, 2009 9:56 am
I think we need to find the number of ways that arrange Frankie and Joey sitting in one row.
xxx 1
xxx 2
Indeed number of possible arrangements is 720 (6!)
So, there are 360 ways in which F&J are sitting in different rows and 360 ways in which they are sitting in the same one.
Am i right ?
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by Brent@GMATPrepNow » Sat Oct 24, 2009 10:12 am
NikolayZ wrote:I think we need to find the number of ways that arrange Frankie and Joey sitting in one row.
xxx 1
xxx 2
Indeed number of possible arrangements is 720 (6!)
So, there are 360 ways in which F&J are sitting in different rows and 360 ways in which they are sitting in the same one.
Am i right ?
I'm not 100% sure if your solution is the same as mine. I'll post mine here as well:

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.

So, the answer is 720/2 or 360.

Extension: If there were n mobsters, then the answer would be n!/2
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by Brent@GMATPrepNow » Sat Oct 24, 2009 10:22 am
Is that similar to your solution dmateer?
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by dmateer25 » Sun Oct 25, 2009 1:20 pm
Brent Hanneson wrote:Is that similar to your solution dmateer?
Yes, my solution was the same as yours!
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by ssuarezo » Thu Oct 29, 2009 7:02 pm
Brent Hanneson wrote:
NikolayZ wrote:I think we need to find the number of ways that arrange Frankie and Joey sitting in one row.
xxx 1
xxx 2
Indeed number of possible arrangements is 720 (6!)
So, there are 360 ways in which F&J are sitting in different rows and 360 ways in which they are sitting in the same one.
Am i right ?
I'm not 100% sure if your solution is the same as mine. I'll post mine here as well:

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.

So, the answer is 720/2 or 360.

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

Sorry, I didn't get it .. why is it clear that Frankie will be standing behind Joey half the time, and Joey will be standing behind Frankie half the time? why half the time?

Thank you,
Silvia.
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by Brent@GMATPrepNow » Thu Oct 29, 2009 8:10 pm
ssuarezo wrote:
Brent Hanneson wrote:
NikolayZ wrote:I think we need to find the number of ways that arrange Frankie and Joey sitting in one row.
xxx 1
xxx 2
Indeed number of possible arrangements is 720 (6!)
So, there are 360 ways in which F&J are sitting in different rows and 360 ways in which they are sitting in the same one.
Am i right ?
I'm not 100% sure if your solution is the same as mine. I'll post mine here as well:

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.

So, the answer is 720/2 or 360.

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

Sorry, I didn't get it .. why is it clear that Frankie will be standing behind Joey half the time, and Joey will be standing behind Frankie half the time? why half the time?

Thank you,
Silvia.
For each arrangement that we can create, we can create another arrangement by switching the positions of Joey and Frankie.
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by gilliamwibson » Thu Mar 11, 2010 5:29 am
think it is a multiplication of two probabilities. AND IT IS NOT ABOUT SITTING , everything happens way before sitting...
one is that Frankie is either in front of the joey or behind that is 1/2 probability that Frankie is going to be behind Joey.
ANd the total probability of arrangements is 720 (5!). Rest is mutliply 1/2 and 720.
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by firdaus117 » Thu Mar 11, 2010 7:44 am
Nature loves symmetry.If she can,why can't we?? :)
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by kstv » Thu Mar 11, 2010 7:55 am
I seems to get a different answer
There are six places to fill.
Joey will not be allowed to stand last so he also has 5 possibilities
seeing him Frankie has 4 places to choose as he cannot be the first in line so he .
so Frankie and Joey can be arranged in 5*4 = 20 ways.
The remaining 4 can be arranged in the 4 vacant slots in 4! ways.
My answer is 20*4! = 480
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by goyalsau » Thu Nov 11, 2010 9:22 am
Brent Hanneson wrote: I'm not 100% sure if your solution is the same as mine. I'll post mine here as well:

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.

So, the answer is 720/2 or 360.

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

In any Combination. Lets say, 5! There are 5 boys,
so any two boys will sit in front or back half the times.......

Can you proof it,, I am not convinced yet, what is the reasoning......................
How can you say sooo...
Saurabh Goyal
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by sdas0112 » Fri Nov 12, 2010 8:27 am
Frankie and Joey need to be together .. treat them as one, so 5 mobsters can be arranged in 5! ways.

Frankie and Joey amongst themselves can be arranged in 2! ways

So 5!*2! = 480
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by Subhadeep2010 » Mon Nov 15, 2010 7:20 am
Simplest way to calculate is the way Brent mentioned i.e 6!/2 ( considering the 2 symmetrical scenarios : Frankie is ahead of Joey , Frankie is behind Joey ).

For those , who are not convinced with this methodology , there is another way.

Let's individually take the scenarios for Joey's postion :-

Joey is 1st => Frankie can now have 5 different positions , rest can have 4! positions . Total = 4!*5
Joey is 2nd => Frankie can now have 4 different positions , rest can have 4! positions . Total = 4!*4
Joey is 3rd => Frankie can now have 3 different positions , rest can have 4! positions . Total = 4!*3
Joey is 4th => Frankie can now have 2 different positions , rest can have 4! positions . Total = 4!*2
Joey is 5th => Frankie can now have 1 different position , rest can have 4! positions . Total = 4!*1

Total possibilties = 4!(5+4+3+2+1)=24*15=360 . Voila , same answer.
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