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Six Mobsters- permutation problem

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by iikarthik » Wed Jul 21, 2010 11:52 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, though not necessarily right behind him. How many ways can the six arrange themselves in line such that Frankie's requirement is satisfied?
6
24
120
360
720

hi,

OA is 360.Pls post explanation
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by Patrick_GMATFix » Wed Jul 21, 2010 2:42 pm
Hi iikarthik,

There are 6 spots. Frankie must be behind Joey. The following are the only possible positions that F and J can take:

If F is in position #1, J can be in 5 different positions (2-6) as follows (assume that position 6 is the front of the line):

(back) F J _ _ _ _ (front)
(back) F _ J _ _ _ (front)
(back) F _ _ J _ _ (front)
(back) F _ _ _ J _ (front)
(back) F _ _ _ _ J (front)

If F is in position #2, J can be in 4 different positions (3-6). Following that logic,
If F is in position #3, J can be in 3 possible positions.
If F is in position #4, J can be in 2 possible positions.
If F is in position #5, J can be in 1 possible position (he must be 1st in line).

So there are 5+4+3+2+1=15 possible ways that F and J can stand.

For each of these 15 ways that F & J can stand, the other 4 mobsters can be arranged in any of 4! arrangements (24 arrangements)

So the total number of possible arrangements is 15*24=360. Pick D.

To create drills with similar questions, set topic='Combinatorics' and difficulty='700+' in the Drill Generator.

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by GMATGuruNY » Wed Jul 21, 2010 5:16 pm
There are 6! = 720 total ways to arrange the 6 mobsters.

Now let's think about this. Isn't the probability that Frankie will be behind Joey the same as the probability that Joey will be behind Frankie? So:

In 1/2 * 720 = 360 of these arrangements, Frankie will be behind Joey.
In 1/2 * 720 = 360 of these arrangements, Joey will be behind Frankie.

Thus, there are 360 ways in which Frankie can be placed behind Joey.

The correct answer is D.
Last edited by GMATGuruNY on Thu Aug 25, 2011 6:58 pm, edited 1 time in total.
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by Patrick_GMATFix » Thu Jul 22, 2010 4:20 am
Classy, elegant solution. Well done and thank you.
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