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Homework Help: Permutation Problem

  1. Dec 14, 2012 #1
    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?

    This is how I tried to solve it. The number of arrangements depend on where Joey stand. So e.g. if Joey is the first in the row, then Frankie has to stand in one of the last four spots of the line so that you have 4! arrangements for those and one of the other 4 mobsters can take the second spot. So the total arrangements for this is 4*4!. Next if Joey is in the second spot, Frankie has to be in one of the last 3 spots, which gives a total of 3*4! arrangements. So in total I got 4!(4+3+2+1)=240 arrangements, however the answer to the question is supposed to be 360. Can anybody tell me what I did wrong?
  2. jcsd
  3. Dec 14, 2012 #2


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    If Joey is first in line, then there are five spots behind him. Not four. Frankie CAN stand right behind Joey, he just doesn't have to.
  4. Dec 14, 2012 #3
    Ah ok thanks a lot. I thought "not necessarily behind him" meant that he is not allowed to stand behind him (not a native speaker).
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