Combinotrics:-)

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Combinotrics:-)

by Shamhoon » Mon Apr 18, 2011 10:33 am
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
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by Stuart@KaplanGMAT » Mon Apr 18, 2011 10:42 am
Shamhoon 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
Hi!

To solve this, let's start with the easier embedded question:

In how many ways can 6 people be arranged in a line?

When we're arranging n distinct objects, there are n! possible arrangements.

So, there are 6! = 6*5*4*3*2 = 720 total arrangements.

(Another way to solve: for the first slot there are 6 possibilities; 5 for the second; 4 for the 3rd; 3 for the 4th; 2 for the 5th; and just 1 for the 6th. So, there are 6*5*4*3*2*1 = 720 arrangements.)

Now let's add in the twist: F must be somewhere behind J.

An important feature of permutations is symmetry. Since we're looking at every possible arrangement, and since F and J are equivalent entities (i.e. there's nothing special about either one), F will be behind J exactly as often as F is ahead of J.

Since F will always be either behind or ahead of J, and since each of those scenarios is equally likely, F will be behind J exactly 50% of the time.

So, to find the total number of arrangements in which F is behind J, we simply take:

total # of arrangements * 1/2 = 720/2 = 360... choose (D)!
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by GMATGuruNY » Mon Apr 18, 2011 10:45 am
Shamhoon 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
There are 6! = 720 total ways to arrange the 6 mobsters.

Now let's think about this. Isn't the probability that Frankie will be behind Joey the same as the probability that Joey will be behind Frankie?

Thus:
In 1/2 * 720 = 360 of these arrangements, Frankie will be behind Joey.
In 1/2 * 720 = 360 of these arrangements, Joey will be behind Frankie.

Thus, the number of ways to place Frankie behind Joey = 360.

The correct answer is D.
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by mk101 » Mon Apr 18, 2011 4:40 pm
This is clearly an arrangement question
For such questions,the constraint applies only to Joey (J) AND Frankie (F).

=> F AND J can be arranged in two ways - FJ , JF.

If we add another person A o F and J the number of arrangements = AJF, AFJ, FAJ,FJA,JFA,JAF.
Thus in the above arrangement carrying F and J we see that the number of times F is ahead of J is equal to the number of times J is ahead of F. ( This rule will apply to any arrangement carrying F and J)

Now coming back to our question - The total arrangements with 6 people = 720

Number of arrangements in which F is ahead of J = Number of arrangements in which J is ahead of F

therefore the required number of arrangements = 720 *1/2 = 360

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by manpsingh87 » Mon Apr 18, 2011 10:55 pm
Shamhoon 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
let joey be represented by j and frankie by f, here we have five possible cases depending upon the position of joey...!!
1) when joey is at the beginning of the row.!
2) when joey is at the second postion of the row.!
3) when joey is at the third position of the row.!
4) when joey is at the fourth position of the row.!
5) when joey is at the fifth position of the row.!

1) j,_,_,_,_,_. now frankie can choose any of the five vacant positions in 5C1 ways, and the remaining four persons can arrange themselves in 4! ways.
2)_,j,_,_,_,_, here frankie can choose any of the four vacant positions behind the joey, in 4C1 ways, and the remaining four persons can arrange themselves in 4! ways.
3) _,_,j_,_,_. here frankie can choose any of the three vacant positions behind the joey in 3C1 ways,
and the remaining four persons can arrange themselves in 4! ways.
4)_,_,_,j,_,_. here frankie can choose any of the two vacant positions behind the joey in 2C1 ways, and the remaining four persons can arrange themselves in 4! ways.
5)_,_,_,_,j,_ here frankie has only one option of picking up the remaining vacant position behind joey, and remaining four persons can arrange themselves in 4! ways.

hence required no. of ways=5*4!+4*4!+3*4!+2*4!+4!,=15*4!=360.!! D
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