Combinatorics 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

Ok, try to understand the numbers: 4!*5+4!*4+4!*3+4!*2+4!=4!*15=360. If still not understandable, let me know, will write more detailed

Thanks. I understand that 4! is necessary to compute the different combinations for the gangsters, other than Joey and Frankie, who are standing in line. However, why are they multiplied by 5, then 4, then 3, then 2, then 1?

We are taking combination's with Joe, namely when the Joe stand the first, second, third and forth
then, if the Joe is the first, there 4! combination and take into account that Frankie can stand in the line 2nd, 3rd, 4th and 5th and 6th - all together 5 - therefore 4!*5
then if Joe is the second, Frankie can be 3rd, 4th and 5th and 6th - 4!*4
and so on, till Joe is the fifth, then frankie can be only the 6th and and the remaining 4 people can vary in the line totaling 4!
so 4!*5 + 4!*4 + ... + 4!*1 = 360

Makes sense, thanks. Always have problems with Combinatorics problems. Lucky for me they're not that common on the GMAT, although I heard recently there are more of them.

total no of possibilities = 6! for all the six people
now half of these will hve joey in front of frankie and other will have frankie in front of joey !!
So answer is simply 6!/2 =360 !