Combinatorics - 700+

This can be attacked in a smilar way as a we do a combinations word problem where in one word needs to be after a specified one. Here F(Frankie) is placed after J(Joey). So if J is to stand 1st among all the 6, F can be placed behind him in 5 ways, rest of the 4 can be arranged in 4! ways. This is one combo.

Next if J is to be placed second among al the 6 in the queue, F can be placed in 5 ways (as F is always behind J ). Remaining 4 can be arranged in 4! ways. This is second combo.

Similarly J can be the 3rd among 6 to stand in the line. So F can be placed in 3 ways..so on.

Finally we have somethign like this:

Total combinations:

5*4!+4*4!+3*4!+2*4!+1*4! = 360.

Hope this explanation helps.

hja379 wrote:Can you please explain? I couldn't understand how to attack this problem.

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

Hi there!

Consider A and B any two of 6 people ordered in a queue.

There are 6! possible ways of arranging all of them. Half this number has A somewhere in front of B, right?

The answer to your problem is therefore 6!/2 = 6*5*4*3 = 30*12 = 360.

Regards,
Fabio.

hja379 wrote:Can you please explain? I couldn't understand how to attack this problem.

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

There are 6! = 720 total ways to arrange the 6 mobsters.

Now let's think about this. In every arrangement, either Frankie will be behind Joey or Joey will be behind Frankie. 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.

So there are 360 ways in which Frankie can be placed behind Joey.

The correct answer is D.

Thank you all so much. Great forum... with smart people !

GMATGuruNY wrote:

hja379 wrote:Can you please explain? I couldn't understand how to attack this problem.

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

There are 6! = 720 total ways to arrange the 6 mobsters.

Now let's think about this. In every arrangement, either Frankie will be behind Joey or Joey will be behind Frankie. 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.

So there are 360 ways in which Frankie can be placed behind Joey.

The correct answer is D.

@ GG - is it possible to solve the same using the slot method (since restrictions are present) ?

Jayanth2689 wrote:

GMATGuruNY wrote:

hja379 wrote:Can you please explain? I couldn't understand how to attack this problem.

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

There are 6! = 720 total ways to arrange the 6 mobsters.

Now let's think about this. In every arrangement, either Frankie will be behind Joey or Joey will be behind Frankie. 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.

So there are 360 ways in which Frankie can be placed behind Joey.

The correct answer is D.

@ GG - is it possible to solve the same using the slot method (since restrictions are present) ?

Joey in Slot 1:
Number of ways to arrange the other 5 mobsters = 5*4*3*2*1 = 120.

Joey in Slot 2:
Since Frankie cannot be in Slot 1, the number of choices for Slot 1 = 4.
Number of ways to arrange the 4 remaining mobsters in Slots 3, 4, 5 and 6 = 4*3*2*1 = 24.
Multiplying, we get 4*24 = 96 arrangements.

Joey in Slot 3:
Since Frankie cannot be in Slots 1 or 2, the number of ways to arrange 2 of the remaining 4 mobsters in Slots 1 and 2 = 4*3 = 12.
Number of ways to arrange the remaining 3 mobsters in Slots 4, 5 and 6 = 3*2*1 = 6.
Multiplying, we get 12*6 = 72 arrangements.

Joey in Slot 4:
Since Frankie cannot be in Slots 1, 2 or 3, the number of ways to arrange 3 of the remaining 4 mobsters in Slots 1, 2 and 3 = 4*3*2 = 24.
Number of ways to arrange the remaining 2 mobsters in Slots 5 and 6 = 2*1 = 2.
Multiplying, we get 24*2 = 48 arrangements.

Joey in Slot 5:
Since Frankie must be in Slot 6, the number of choices for Slot 6 = 1.
The number of ways to arrange the remaining 4 mobsters in Slots 1, 2, 3, and 4 = 4*3*2*1 = 24.
Multiplying, we get 1*24 = 24 arrangements.

Adding the results above, we get:
120+96+72+48+24 = 360 arrangements.

If there were no restrictions then we could have 6! = 720 ways

If we have Frankie in the first slot we will have 5! = 120 ways

But Frankie can be in other slots as well. So the only answer 120 < 360 < 720 should be the solution.

IMO D

GMATGuruNY wrote:

Jayanth2689 wrote:

GMATGuruNY wrote:

hja379 wrote:Can you please explain? I couldn't understand how to attack this problem.

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

There are 6! = 720 total ways to arrange the 6 mobsters.

Now let's think about this. In every arrangement, either Frankie will be behind Joey or Joey will be behind Frankie. 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.

So there are 360 ways in which Frankie can be placed behind Joey.

The correct answer is D.

@ GG - is it possible to solve the same using the slot method (since restrictions are present) ?

Joey in Slot 1:
Number of ways to arrange the other 5 mobsters = 5*4*3*2*1 = 120.

Joey in Slot 2:
Since Frankie cannot be in Slot 1, the number of choices for Slot 1 = 4.
Number of ways to arrange the 4 remaining mobsters in Slots 3, 4, 5 and 6 = 4*3*2*1 = 24.
Multiplying, we get 4*24 = 96 arrangements.

Joey in Slot 3:
Since Frankie cannot be in Slots 1 or 2, the number of ways to arrange 2 of the remaining 4 mobsters in Slots 1 and 2 = 4*3 = 12.
Number of ways to arrange the remaining 3 mobsters in Slots 4, 5 and 6 = 3*2*1 = 6.
Multiplying, we get 12*6 = 72 arrangements.

Joey in Slot 4:
Since Frankie cannot be in Slots 1, 2 or 3, the number of ways to arrange 3 of the remaining 4 mobsters in Slots 1, 2 and 3 = 4*3*2 = 24.
Number of ways to arrange the remaining 2 mobsters in Slots 5 and 6 = 2*1 = 2.
Multiplying, we get 24*2 = 48 arrangements.

Joey in Slot 5:
Since Frankie must be in Slot 6, the number of choices for Slot 6 = 1.
The number of ways to arrange the remaining 4 mobsters in Slots 1, 2, 3, and 4 = 4*3*2*1 = 24.
Multiplying, we get 1*24 = 24 arrangements.

Adding the results above, we get:
120+96+72+48+24 = 360 arrangements.

Excellent!!! i was majorly confused with Joey's placement when i tried the slot method! Thanks GG!