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
I am little confused with the explanation given by MGMAT.
Suppose, I take the scenario like this -
1 2 3 4 5 6 -> 6 positions and A B C D J F -> 6 mobsters.
Now
If J is at position - 1, F can be in other 5 positions.
If J is at postion - 2, F can be in other 4 positions. [After J]
If J is at postion - 3, F can be in other 3 positions. [After J]
If J is at postion - 4, F can be in other 2 positions. [After J]
If J is at postion - 5, F can be in other 1 positions. [After J]
If J is at position - 6, F can not stand after that -> So we can consider the case that J can not stand at 6 position.
So the different combination is = 5 * 4 * 3 * 2 * 1 = 120.
So go for C. But why the OA is 360? Am I missing something ...