Came across this question:
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?
Ans:
6
24
120
360
120
Explanation given was:
Ignoring Frankie's requirement for a moment, observe that the six mobsters can be arranged 6! or 6 x 5 x 4 x 3 x 2 x 1 = 720 different ways in the concession stand line. In each of those 720 arrangements, Frankie must be either ahead of or behind Joey. Logically, since the combinations favor neither Frankie nor Joey, each would be behind the other in precisely half of the arrangements. Therefore, in order to satisfy Frankie's requirement, the six mobsters could be arranged in 720/2 = 360 different ways.
The correct answer is D.
I am solving it in other way.
there are 6 possible places -1 -2 -3 -4 -5 -6
if J is at 1st : F can be at 5 possible places
if J is at 2nd : F can be at 4possible places
if J is at 3rd : F can be at 3possible places
if J is at 4th : F can be at 2possible places
if J is at 5th: F can be at 1possible places
Multiplication rule of counting:
5*4*3*2*1 = 120 places.
Can u point me where I am going wrong...this is the most general step I follow for solving all permutation problems and it has never failed.
Regards,
Bhawna