Anybody please help me to explain more about this problem : MGMAT CAT1

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. 6
B. 24
C. 120
D. 360
E. 720

To permutate 6 people such as " a b c d e f " with "a" and "b" together so I replace "a" and "b" by "A". The problem now becomes " A c d e f " => I have 5! = 120 ways to arrange them then multiple by 2 ( because "A" = "ab" or "ba" ) so 120 * 2 = 240 (ways).

Why the answer is 360. Please help me to make it clear ?

Your solution works if Frankie is DIRECTLY behind Joey, since you paired them up to be one. However, Frankie can be ANYWHERE behind Joey. For example, if Frankie is F, Joey is J, and the other mobsters are A, B, C, and D, these are possible arrangements, from back to front:
J, F, C, D, F, J
J, A, B, C, D, F
A, J, B, F, C, D
A, J, B, C, F, D

So in order to solve this problem, we should figure out all the positions Joey can be in line. There are FIVE possible positions:
(1) J, X, X, X, X, X
(2) X, J, X, X, X, X
(3) X, X, J, X, X, X
(4) X, X, X, J, X, X
(5) X, X, X, X, J, X

Let's take position (1). In this case, it doesn't matter which position Frankie takes, he will always be behind Joey. If we use the slot approach, it would be:
J, X, X, X, X, X
(1)(5)(4)(3)(2)(1) = 120

For position (2), Frankie has to be in the four slots behind J, and the slot in front of J can be A, B, C, or D (so there are four options there). After the first slot is taken, there four remaining people to fill up the four remaining slots (Frankie and the other three not selected).

X, J, X, X, X, X
(4)(1)(4)(3)(2)(1) = 96

If you use the same reasoning on position (3), we have:
X, X, J, X, X, X
(4)(3)(1)(3)(2)(1) = 72

For position (4):
X, X, X, J, X, X
(4)(3)(2)(1)(2)(1) = 48

Finally, for position (5):
X, X, X, X, J, X
(4)(3)(2)(1)(1)(1) = 24

If you add up all the combinations: 120 + 96 + 72 + 48 + 24 = 360.

Hope that helps.

_________________
Ben Ku
Instructor
ManhattanGMAT

Thank BenKu,

Your explanation is very clear and detail. I'm really appreciated !
It fell to my silly mistake. I usually have problems with the long & complicated topic.

Glad it helped.

_________________
Ben Ku
Instructor
ManhattanGMAT

Put Joey on place 1, franky has 5 place to select now, behind him. And rest 4 can select any other place so - 4!
Joey on Place 2, franky goes for 4 positions. And rest can go for 4!

Similarly Joey goes till 5th position and Franky for 1 and rest will remain the same with 4!

Summing all these up
4! (5+4+3+2+1) = 24x15 = 360

As with many combinations problems, this one can be solved through a much more intuitive approach:
There are 720 ways to line up the guys.
In every one of these, Frankie is either ahead of Joey or behind him.
Because 720 represents all possible orderings, exactly half of them will have Frankie ahead of Joey and the other half will have Joey ahead of Frankie.
720/2=360.

_________________
Tim Sanders
Manhattan GMAT Instructor