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

This is a question about permutations: how many ways can the six mobsters form a line in such a way that Joey comes before Frankie. I solved a similar problem using brute force in a thread asking about combinations. That problem involved five people... six people is starting to push the brute force envelope, so I'll try a different method here:

First we need to know all the ways Frankie and Joey can be arranged in the line
Joey 1st - Frankie can be in any of the 5 remaining places: 5 arrangements
Joey 2nd - Frankie can be in any of the 4 places behind Joey: 4 arrangements (see a pattern?)
Joey 3rd - Frankie can be in any of the 3 places behind Joey: 3 arrangements (now, do you see the pattern?)

The number of possible arrangements of Joey followed by Frankie is 5 + 4 + 3 + 2 + 1 = 15.

Now consider the other mobsters. The 4 other guys can go in the four remaining spots of any of these arrangements in any order. The number of possible arrangements of the four other goodbuddies is 4*3*2*1 = 24.

So the total number of arrangements is 24*15 = 360

BenchPrepGURU
benchprep.com

I posted a solution here:

https://www.beatthegmat.com/six-mobsters ... tml#276472