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Counting - Six mobsters

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by Brent@GMATPrepNow » Tue Jan 31, 2012 7:30 am
jsnipes wrote:i got the right answer but i think i may have done it wrong. so total number of ways that the 6 people can be arranged (since order matters?) is 6! or 720. i then took 720 and divided it by the number of people who have to stand together as 2! not two. so i got 6!/2! which for 2 is obviously the same thing and results in the correct answer of 360. if there were three people would I divide 720 by 3 or 3!

thanks for help, i know this is two day old question i just got around to answering it though.
Great question.
If there were three people to consider (A, B and C) and we wanted to know how many 6-person configurations are such that A is ahead of B, and B is ahead of C, then we'd divide 720 by 3!

We'd divide by 3! since there are 3! (6) ways to arrange A, B and C and only one of those 6 arrangements would be such that A is ahead of B, and B is ahead of C.

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by sanjila007 » Wed Jun 13, 2012 10:14 am
Brent@GMATPrepNow 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:

Hey Brent cant we proceed that taking F&J as same quantities then there would be 5! ways to arrange them in a row but then I am not able to think.Please proceed with this approach if it is true...

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 tisrar02 » Thu Jun 14, 2012 6:35 am
I guess you can also think of this question as when ORDER MATTERS. So we have 6 random guys so we have 6!. But we want it so that Joey is ALWAYS in front of Frankie. So when dealing with permutations in terms where order does matter, you would go n!/(n-r)!... N= total number of people and r would be the remaining bunch other than the 2 that we want to know about.

6!/(6-4)!= 720/2= 360 different ways to how they can be set up.



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by gmattest001 » Thu Jun 14, 2012 11:50 pm
Great solution , but need time to digest.

Brent@GMATPrepNow 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
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by Ganesh hatwar » Fri Jul 06, 2012 1:21 am
Frankie is behind Jack

if J is in first place there are 5 possibilities for Frankie.

and If Jackie is in 2 place then 4 possibilities for Frankie.

same way Jackie in 3rd , 2nd and First 1 .. 3,2 and 1 possibility.

so total of 15 15 possibilities and 4 other can arrange themselves in 4! ways

so 15*4*3*2*1 = 360 !!

took 4.30 minutes!!!

PS : An awesome question
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by cuddytime » Mon Jul 23, 2012 9:55 am
took < 30 seconds for me.

We know the smallest restriction is F&J together so that would be: 5!
We know the largest restriction (No line restriction) is: 6!

# has to be somewhere in between so: D.
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by mba404 » Sun Aug 12, 2012 10:01 pm
Total number of ways = 6! = 6 * 5! = 720 (I just remember that 5! = 120. so just used it as a shortcut to calculate 6!)

Arrangerment1 = F standing back of J
Arrangerment2 = J Standing back of F

These two are mutually exclusive and when added give the total number of ways. (There can't be an arrangement where F&J are standing in the same position)

ways(Toatl) = ways( Arrangerment1 ) + ways( Arrangerment2 )

ways(Arrangerment1) = ways( Arrangerment2 ) = x -- as each is an arrangement where we want one guy to be in front of some other guy.

720 = x + x
x = 720/2 = 360 (D)



Total number of ways
Required number of arrangements would be exactly 1/2 of total number of arrangements.
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by rajeshsinghgmat » Tue Jan 22, 2013 8:22 pm
I D(id) it.

Total no. of arrangements = 6!

In one half of the arrangements Frankie will be behind Joey and in the other half Frankie will be before Joey.

There for no. of required arrangements = (1/2)(6!) = 360
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by vongochao » Mon Feb 11, 2013 7:59 am
the question is here ? how many lines? if one line is so easy to do, otherwise! teeheee.teeheee.......
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by vishalbpr » Tue Feb 12, 2013 2:10 am
Brent@GMATPrepNow 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
Awesome answer. It was out of the way thinking. If I face such questions in exam, now I can think in a different way.
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by jiarossi » Mon Feb 18, 2013 9:05 am
I think my confusion with this problem is around the question wording. The part that says "insists upon standing behind Joey in line at the concession stand, though not necessarily right behind him." makes me think there must be a space between joey and frankie. I read the problem as though if Joey was in spot one then Frankie could not be in spot two. His first available choice would be spot 3.

How do you know that is not what the question is looking for? I was really tricked up with this.
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by Brent@GMATPrepNow » Mon Feb 18, 2013 9:19 am
jiarossi wrote:I think my confusion with this problem is around the question wording. The part that says "insists upon standing behind Joey in line at the concession stand, though not necessarily right behind him." makes me think there must be a space between joey and frankie. I read the problem as though if Joey was in spot one then Frankie could not be in spot two. His first available choice would be spot 3.

How do you know that is not what the question is looking for? I was really tricked up with this.
I think you're equating "not necessarily" with "not"

Let's examine two instances where the Official Guide uses "not necessarily"

Figures are not necessarily drawn to scale.
This isn't saying that the figures are not drawn to scale. All it's saying is that we cannot assume that they are drawn to scale. It's still possible that a figure is drawn to scale.

Manufacturing a premium brand is not necessarily more costly than manufacturing a standard brand of the same product
In other words, we cannot assume that manufacturing a premium brand is more costly than manufacturing a standard brand. That said, it is still possible that manufacturing a premium brand is more costly than manufacturing a standard brand. This not mean, however, that manufacturing a premium brand is less costly than manufacturing a standard brand.

So, with the original question, we have: Frankie is behind Joey, though not necessarily right behind him.
Here, we cannot assume that Frankie is directly behind Joey, although it is possible.

I hope that helps.

Cheers,
Brent
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by petrifiedbutstanding » Tue May 28, 2013 11:52 pm
Brent@GMATPrepNow wrote: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
Brett, what about the condition here that says "not necessarily right behind him"?

I took that into consideration and got 24.
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by petrifiedbutstanding » Tue May 28, 2013 11:56 pm
I see you've already answered that. So is this a rule of thumb when you notice "not necessarily" on the GMAT?
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by Brent@GMATPrepNow » Wed May 29, 2013 5:23 am
petrifiedbutstanding wrote:I see you've already answered that. So is this a rule of thumb when you notice "not necessarily" on the GMAT?
That's correct. In fact, it also applies to instances that are not on the GMAT.

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