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. 6b. 24c. 120d. 360e. 720

The correct answer is D) 360 possible arragements.

Step-by-step explanation:

To solve the problem, we use the slot method. There are 6 possibilities:

Possibility N°1: J _ _ _ _ _  (Joey is first in the line)

Possibility N°2: _ J _ _ _ _  (Joey is second in the line)

Possibility N°3: _ _ J _ _ _  (Joey is third in the line)

Possibility N°4: _ _ _ J _ _  (Joey is fourth in the line)

Possibility N°5: _ _ _ _ J _  (Joey is fifth in the line)

Possibility N°6: _ _ _ _ _ J (Joey is sixth in the line)

Frankie must always stand behind Joey.

Possibility N°1: there are 5! possible arragements , 5!=120

Possibility N°2: Frankie can't be ahead of Joe, so possible cases, 5! - 4! = 120 -24 = 96

Possibility N°3: Frankie can't take be at the first two positions, so 5! - 2×4! = 120 -48 = 72

Possibility N°4: Frankie can't take the at the first three, so 5! - 3×4! = 120 - 72 = 48

Possibility N°5: Frankie can't take be at first 4, so 5! - 4×4! = 120-96 = 24

Possibility N°6: It does not satisfy the criteria.

The sum the possibilities is 120+96+72+48+24=360 possible arragements.

i can't see the picture : /