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MGMAT permutations question.

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by ratindasgupta » Wed Sep 26, 2007 10:47 am
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
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by gmatguy16 » Wed Sep 26, 2007 11:11 am
imo 360
4! + 2*4! + 3* 4$ + 4*4! +5* 4! = 360 ...
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by givemeanid » Wed Sep 26, 2007 11:14 am
6!/2 = 720/2 = 360
So It Goes
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by gmatguy16 » Wed Sep 26, 2007 11:15 am
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 ...
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by kajcha » Wed Sep 26, 2007 11:17 am
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
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by ratindasgupta » Wed Sep 26, 2007 12:56 pm
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! ?
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by ratindasgupta » Wed Sep 26, 2007 1:03 pm
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! ?
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by kajcha » Wed Sep 26, 2007 1:07 pm
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
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by ratindasgupta » Wed Sep 26, 2007 1:23 pm
Thanks kajcha!

Got it. (kicking myself for making such a fundamental mistake)
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by persevering » Wed Sep 26, 2007 2:44 pm
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!
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