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Number of ways to line up, with restrictions

  1. Apr 23, 2009 #1
    1. The problem statement, all variables and given/known data
    Seven married couples, five of which with a single child, go to a movie theater. In how many ways can they line up to buy tickets so that no two children stand next to each other?

    This is the problem, word for word.
    The problem does not state that families have to be together. The part (b), which I have done already, did specifically say families are together. So, this is being done on the assumption that since it was not stated families are together, they do NOT have to be together.

    2. Relevant equations
    Arranging them without restriction = 19!
    choosing combos: n!/(r!)(n-r)!


    3. The attempt at a solution
    Here is my thought process:
    10 slots in the line have to be filled by a pair of child/adult. So I did 19!/(10!)(9!). That gives the number of different ways that those slots can be filled. Then there are 9 remaining slots that are filled by 9 adults, so I did 9!. So the whole thing would be:
    19!/(10!)(9!) * 9! = 19!/10!

    I just have a feeling that thats not right.
     
    Last edited: Apr 23, 2009
  2. jcsd
  3. Apr 23, 2009 #2
    Should you assume each family stands together, or is it ok for spouses to stand way apart? If way apart, should a kid stand with one or both parents, or can the kid stand way apart also? Since they are buying tickets, I would tend to assume the family stands together.
     
  4. Apr 23, 2009 #3
    Edited my first post. The problem doesn't state that they have to stay together, so assume they don't. Our professor who made the problem is out of town right now, and will be until after it is due, so I can't ask him for clarification.
     
  5. Apr 23, 2009 #4
    No two children should be together, right?

    Consider two cases.

    1) Children occupy even places. Now we have 19 places of which 9 are even. We have to arrange 5 children in such 9 places(which you would be knowing). Doing so will leave 4 even places and 10 odd places unoccupied in which you have to arrange the couples(Which you would be knowing).

    2)Children occupy odd places. 10 are odd. Arrabge the five children in these ten places. Abd follow as the first case.
     
  6. Apr 24, 2009 #5
    But then there could be a child in an even space, then two adults, then a child in an odd space. Wouldn't that mess things up?
     
  7. Apr 24, 2009 #6
    Perfect reasoning. So lets try the correct way.

    we have got 14 adults which can be arranged in 14! these 14 people create 15 places. Fill up this 15 places by 5 children.
     
  8. Apr 24, 2009 #7
    Oh hey that makes sense.

    So maybe (14! * (15 choose 5) * 5!) ?
     
  9. Apr 24, 2009 #8
    It would be just 14!*(15C5) You don't have to multiply with 5! cause we already selected five places out of 15(its just like arranging them) so you dont have to then multiply with 5!
     
  10. Apr 24, 2009 #9
    Alright, thank you so much!
     
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