I am not good with permutation & combinations. But, I will try.

The requirement is that Frankie is always behind Joey, not necessarily immediately behind him. That means Joey cannot be the last person in the line (because Frankie should be behind him). So, Joey can be placed in any of the first five positions in 5 ways.

When Joey is in the
I position, Frankie has 5 ways of standing in the line & the rest can stand in 4! ways, i.e. 5X4X3X2X1 = 120 ways
II position, Frankie has 4 ways of standing in the line & the rest can stand in 4! ways, i.e. 4X4X3X2X1 = 96 ways
III position, Frankie has 3 ways of standing in the line & the rest can stand in 4! ways, i.e. 3X4X3X2X1 = 72 ways
IV position, Frankie has 2 ways of standing in the line & the rest can stand in 4! ways, i.e. 2X4X3X2X1 = 48 ways
V position, Frankie has 1 way of standing in the line & the rest can stand in 4! ways, i.e. 1X4X3X2X1 = 24 ways

Adding all the possibilities = 120+96+72+48+24 = 360ways.

I hope my approach & answer are correct

OA is D


Vemuri ur approach is nice.. i was actually looking for this kind of an explanation bcos according to the wording of the problem i interpreted it as frankie shud always be behind joey as frankie wants to keep an eye on joey. the way the test takers have explained it they have concidered tht frankie can be ahead or behind joey and thus halfed the total combination ie 6! /2.

I think ur explanation is the best tht i can understand.[/i]