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?

answer choices are
a.6
b.24
c.120
d.360
e.720

I chose C, i thought frankie and joey can be considered one unit, now the combinations of 5! is 120. Apparently, i should have multiplied 5!*2 because frankie can be infront of joey or behid joey.

Answer is 360. I could not understand the explanation. Please help.

There are a couple of ways for doing this:

1.

Think of 6 spots:

1. F chooses place #1, other 5 can be arranged in 5! Ways
2. F chooses place #2, J has to choose one of 4, and other 4 guys can be arranged in 4! Ways
3. F chooses place #3, J has to choose one of 3 behind, and other 4 guys – 4!
4. …

5! + 4 * 4! + 3 * 4! + 2 * 4! + 1 * 4!
5! + 4!(4 + 3 + 2+ 1)
5! + 4!(10) = 360

2.

MGMAT CAT explanation (which is much nicer): 6! ways to arrange all 6 = 720.
Logically in half of them J would be behind F, so 720/2 = 360 :)

Thanks TP!

MGMAT explanation is simple. May be, i started to become a linear lover :-)