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Permutations seat arrangements

  1. Mar 4, 2009 #1
    1. The problem statement, all variables and given/known data

    9 people, 2 cars can hold 5 people each. only 3 people have licenses.

    2. Relevant equations

    How many different ways can the people be seated into cars?

    3. The attempt at a solution

    There are nine people. So number of seats is not a problem.

    We must find out how many groups of three people with licenses there can be.

    nPr = n!(n-r)!
    n = 9
    r = 3
    nPr= 9!/(9-3)! = 504


    How's that look?
     
  2. jcsd
  3. Mar 4, 2009 #2

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Re: Permutations

    How did you arrive at that? The problem is to divide the 9 people into 2 groups. And, since there are only two cars, in each possibility one of the people with a license is irrelevant.

     
    Last edited: Mar 5, 2009
  4. Mar 4, 2009 #3
    Re: Permutations

    So It works out to be

    nPr = n!(n-r)!
    n = 4
    r = 3
    nPr= 4!/(1)! = 24 +


    nPr = n!(n-r)!
    n = 4
    r = 3
    nPr= 4!/(1)! = 24


    = 48?
     
  5. Mar 5, 2009 #4
    Re: Permutations

    Just been looking at this problem again.

    If there are 3 licensed drivers isn't there 3! (six times) they can be arranged ?

    Which would leave 6 pasengers arranged into two cars say

    3 in car A (+ 2 drivers) and 3 in car B (+ 1 driver) = nine all up

    Would you than calculate:

    3!(6C3) + 3!(6C3)

    nCr = n!(n-r)!r!
    > n = 6
    > r = 3
    > nCr= 6!/(3)!3!
    >
    > =20 * 3!
    =120
    >
    +
    >
    > nCr = n!(n-r)!r!
    > n = 6
    > r = 3
    > nCr= 6!/(3)!3!
    = 20 * 3!
    = 120


    = 240 ?

    This is very confusing AAARRRHHH !

    Thanks,
    Brendan
     
  6. Mar 5, 2009 #5

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Re: Permutations

    If you say "and then take the first two to be drivers of the two cars (in order)" then, yes, that is correct.

    Be careful. There are not three cars so one of the people with a license has to be included amoung the 7 passengers.


    You are assuming that the licensed person who is not driving will be in car A. That is not given.

     
  7. Mar 5, 2009 #6
    Re: Permutations

    I see what you mean.

    There are 3! drivers = 6
    But after the 2 drivers are selected there are still 7 people to be allocated.

    Therefore there are 7C4 = 35 combinations of people.

    So we multiply the number of passenger combinations by the driver combinations to get

    35 X 6 = 210.

    Personally I would tell them all just to get in however they wanted!
    regards
    Brendan
     
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