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.
