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Variations question

  1. Feb 16, 2008 #1
    1. The problem statement, all variables and given/known data

    On how many ways, 8 different people can sit on 12 seats?

    2. Relevant equations

    [tex]V\stackrel{k}{n}[/tex]

    3. The attempt at a solution

    I tried to solve this task by using the formula of variations without repetition.

    [tex]V\stackrel{8}{12}[/tex]

    But I sow in my textbook results, and it says that it should be: 8! * [tex]V\stackrel{8}{12}[/tex]

    I don't know what is correct actually.
     
  2. jcsd
  3. Feb 16, 2008 #2

    HallsofIvy

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    I have absolutely no idea what [itex]V_n^k[/itex] means!

    I suspect that that it is what I would call [itex]_nC_k[/itex] or [itex]\left(\begin{array}{c}n \\ k \end{array}\right)[/itex], the binomial coefficient. That is the number of ways we can pick a group of 8 chairs out of 12 for the people to sit on. That, however, does not include the number of orders in which those 8 people can sit on the same 8 chairs. Since there are 8! orders for 8 people, that would be [itex]8!_12C_8[/itex] or 12!/(12-8)!= 12!/4!
     
  4. Feb 17, 2008 #3
    [itex]V_n^k[/itex]=n(n-1)(n-2)...(n-k)
    So, 8!*[itex]V_12^8[/itex] will come to be, big number.
    8! * 12*11*10*9*8*7*6*5
     
  5. Feb 17, 2008 #4

    HallsofIvy

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    That is almost what I would call the number of Permutations:
    [tex]_nP_k= \frac{n!}{(n-k)!}[/tex]
    ("Almost" because what you have is
    [tex]\frac{n1}{(n-k-1)!}[/tex])
    and that is exactly what I gave before: k! nCk. If you text really has another 8! I have no idea why they are doing that. Perhaps it is a typo.
     
  6. Feb 17, 2008 #5
    I also hope so, that it is typo. Because if I get 2 people to sit on 3 seats (simplified edition of the task above), there will be 6 ways to sit.
    [itex]V_3^2[/itex]=3*2=6
    or
    2!*[itex]C_3^2[/itex]=2*3=6
     
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