manhattan gmat combinatorics

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?
1) 6
2) 24
3) 120
4) 360
5) 720

I do not understand the solution provided by MGMAT where they mention that "each of those 720 arrangements, Frankie must be either ahead of or behind Joey". Now they have this constraint that Frankie will always be behind Joey so how come he must be either ahead or behind Joey.

Can someone please explain this in some other way.

Thanks.

MGMAT explanation

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.