Problem Solving

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?


6
24
120
360
720

is te answer 360

the answer is 360 can u explain how u solved it .

any other method used by anyone?

any other method used by anyone?

younggun044,

You have to fix Joey in all the possible positions and calculate the possible arrangements for the remaining people.

1st: Joey is on the first position

Therefore no matter what is the position of Frankie he'll be behind Joey. Since Joey is fixed, the remaining five have 5! ways of being arranged.

2nd: Joey is in the second position

Frankie must be on position nr. 3, 4, 5 or 6, meaning 4 possible positions. The remaining 4 people must be arranged in the remaining 4 positions. So 4*4!

3rd: Joey is in third position

Frankie must be on position nr. 4, 5 or 6, meaning 3 possible positions. The remaining 4 people must be arranges in the remaining 4 positions. So, 3*4!

4th: Joey is in fourth position

Frankie must be on position nr. 5 or 6, 2 possible positions. The remaining 4 people can be arranged in the 4 remaining positions. So 2*4!

5th: Joey is in fifth position

Frankie must be on position nr. 6, one possible position. The remaining 4 people can be arranged in the remaining 4 positions, so 4!

Joey cannot be on position nr. 6 because this is the last position, meaning that Frankie could not be behind him.

So, 5! + 4*4! + 3*4! + 2*4! + 4! = 360

Hope it helps.

yup...ofcourse it does!!