MGMAT permutations question.

[ratindasgupta]

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. 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

[gmatguy16]

imo 360
4! + 2*4! + 3* 4$ + 4*4! +5* 4! = 360 ...

[givemeanid]

6!/2 = 720/2 = 360

[gmatguy16]

sorry its not $ its ! (factorial)
one trick ..if there is no restriction then there are 6 ! = 720 ways
if the restriction is that frank and joey need to be immediately behind each other then there are 120 ways ...
and our case is in between ...

[kajcha]

Ans should be 360.

6 people can be arranged in 6! = 720 ways

Half of this combinations J will be in front of F.

So total combinations = 720/2 = 360

[ratindasgupta]

the answer is 360.

I'm a bit confused as i'm getting the answer as 120. if we consider F&J as one, since they're always together, shouldn't the answer be 5! ?

[ratindasgupta]

the answer is 360.

I'm a bit confused as i'm getting the answer as 120. if we consider F&J as one, since they're always together, shouldn't the answer be 5! ?

[kajcha]

ratin, If you consider F&J as one unit you are calculating when J is just in front of F. But the question is asking something different. e.g. F can be at 6th position and J can fill any of 5 positions in front of him. From your calculation J is filling only 5th position.

Hope this clears your doubt

[ratindasgupta]

Thanks kajcha!

Got it. (kicking myself for making such a fundamental mistake)

[persevering]

drawing to illustrate the answer by gmatguy16:

xxxxxF (J=5) J has 5 places
xxxxFx (J=4) J has 4 places
xxxFxx (J=3)
xxFxxx (J=2)
xFxxxx (J=1)

sum over (J * 4! ) = (1+2+..5) * 4!