Combinatorics question from MGMAT CAT

[Mr Muggles]

I did my first MGMAT CAT and got the following question:

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

Here is how I approached it:
- if Joey is in the first car, Frankie has 5 possible spots
- if Joey is in the 2nd car, Frankie has 4 possible spots (since he won't be in front of Joey)
etc
so answer = 5x4x3x2 = 120.

WRONG!
Here is the official answer from MGMAT:

Ignoring Frankie's requirement for a moment, observe that the six mobsters can be arranged 6! or 6 x 5 x 4 x 3 x 2 x 1 = 720 different ways in the concession stand line. In each of those 720 arrangements, Frankie must be either ahead of or behind Joey. Logically, since the combinations favor neither Frankie nor Joey, each would be behind the other in precisely half of the arrangements. Therefore, in order to satisfy Frankie's requirement, the six mobsters could be arranged in 720/2 = 360 different ways.

The correct answer is D.

Can anyone explain this to me please? What is wrong with my approach?

[gaggleofgirls]

I think the problem with your approach is that you aren't taking into account the order of the other 4 people.

For this question, J F A B C D is not the same as J F B C D A.

-Carrie

[krisraam]

I did my first MGMAT CAT and got the following question:

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

Here is how I approached it:
- if Joey is in the first car, Frankie has 5 possible spots
- if Joey is in the 2nd car, Frankie has 4 possible spots (since he won't be in front of Joey)
etc
so answer = 5x4x3x2 = 120.

WRONG!
Here is the official answer from MGMAT:

Ignoring Frankie's requirement for a moment, observe that the six mobsters can be arranged 6! or 6 x 5 x 4 x 3 x 2 x 1 = 720 different ways in the concession stand line. In each of those 720 arrangements, Frankie must be either ahead of or behind Joey. Logically, since the combinations favor neither Frankie nor Joey, each would be behind the other in precisely half of the arrangements. Therefore, in order to satisfy Frankie's requirement, the six mobsters could be arranged in 720/2 = 360 different ways.

The correct answer is D.

Can anyone explain this to me please? What is wrong with my approach?

Your approach has Joe and Frankie always together. The requirement is Joe has to be ahead of frankie in the line. They doesn't need to be together.

F_ _ _ _ _ = 4! * 5( Joe can in any of the 5 empty spots)
_ F _ _ _ _ = 4! * 4( Joe can in any of the 4 empty spots)
_ _ F _ _ _ = 4! * 3( Joe can in any of the 3 empty spots)
_ _ _ F _ _ = 4! * 2 ( Joe can in any of the 2 empty spots)
_ _ _ _ F_ = 4! * 1( Joe can in any of the 1 empty spots)

Total no of ways = 4!( 5 + 4 + 3 + 2 + 1) = 360.

Thanks
Raama

[Mr Muggles]

thanks guys, now I get it