Permutation Question

[jsloan01]

I need some help with digesting the answer to 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?

The answer:

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.

My thoughts:

I understand the 6! There are 720 arrangements without the constraint. However, I don't understand why we divide by 2. I would think that since Frankie wants to stand behind Joey, the limitations would be:
Slot 1: Joey, Frankie has 5 choices
Slot 2: Joey, Frankie has 4 choices
Slot 3: Joey, Frankie has 3 choices
and so on.

Thanks in advance

[kundan_scorpio]

Hi,
I would approach this problem this way:
if there are 6 people to be arranged in 6 places, then number of ways they can be arranged would be 6! ways.

Since Frankie has to be always behind Joe not necessarily immediately behind him:
So let us fix Joe's position & count the number of options for frankie & remaining people:
A : if Joe at 1st position then Frankie has 5 slots behind him.
B
C
D
E
F

Say for eg: he occupies 3rd slot then remaining 4 persons can be arranged in total 4! ways. This means no of ways this option will have is
a) 1(for Joe) X 5 (for Frankin) X 4!(for remaining 4)

Second possibility:
A
B Joe's position
C
D
E
F

Now Joe is at second position so franlin has 4 slots (if he has to remain behind him)
b) This means no of ways this option will have is
1(for Joe) X 4 (for Frankin) X 4!(for remaining 4)
Similarly
c) 1(for Joe) X 3 (for Frankin) X 4!(for remaining 4)
d) 1(for Joe) X 2 (for Frankin) X 4!(for remaining 4)
e) 1(for Joe) X 1 (for Frankin) X 4!(for remaining 4)

Add all the options from a to e: You will get 360.
Hope this helps.

Cheers
Kundan

[jsloan01]

Hi,
I would approach this problem this way:
if there are 6 people to be arranged in 6 places, then number of ways they can be arranged would be 6! ways.

Since Frankie has to be always behind Joe not necessarily immediately behind him:
So let us fix Joe's position & count the number of options for frankie & remaining people:
A : if Joe at 1st position then Frankie has 5 slots behind him.
B
C
D
E
F

Say for eg: he occupies 3rd slot then remaining 4 persons can be arranged in total 4! ways. This means no of ways this option will have is
a) 1(for Joe) X 5 (for Frankin) X 4!(for remaining 4)

Second possibility:
A
B Joe's position
C
D
E
F

Now Joe is at second position so franlin has 4 slots (if he has to remain behind him)
b) This means no of ways this option will have is
1(for Joe) X 4 (for Frankin) X 4!(for remaining 4)
Similarly
c) 1(for Joe) X 3 (for Frankin) X 4!(for remaining 4)
d) 1(for Joe) X 2 (for Frankin) X 4!(for remaining 4)
e) 1(for Joe) X 1 (for Frankin) X 4!(for remaining 4)

Add all the options from a to e: You will get 360.
Hope this helps.

Cheers
Kundan

Awesome reply. Thank you!

[ashk29]

F and J cannot stand in one place , because of that either J is in front of F or behind F

Total no. of possibilities = 720 , but half the time F is behind J ( the question specifically says that F need not be just behind J)
so ans= 720/2 = 360

[mightyalworth]

Question: if I modify the sentence "though not necessarily right behind him" with "not right behind him", what would be the result? n!/2n ?