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[Probability] Seating arrangement & Selection

  1. Nov 20, 2009 #1
    PROBLEM 1
    7 people sit in a row, and 2 of them want to sit together, in how many ways can they be arranged?

    ATTEMPTED SOLUTION:
    6! x 2 = 1440
    -------------------------------------------------------------------------------------------------------
    PROBLEM 2
    7 people sit in a row, and 2 of them do not want to sit together, in how many ways can they be arranged?

    ATTEMPTED SOLUTION:
    No ideas
    -------------------------------------------------------------------------------------------------------
    PROBLEM 3
    7 people sit in a circular formation, but 2 of them do NOT want to sit together, in how many ways can they be arranged?

    ATTEMPTED SOLUTION
    No ideas
    -------------------------------------------------------------------------------------------------------
    PROBLEM 4
    There are 10 red & 6 blue marbles in a bag. You pick 2 of them at the same time.
    In how many ways can you pick 2 marbles of the same color? Different colors?

    ATTEMPTED SOLUTION:
    Same colors: 10/16 x 9/15 + 6/16 x 5/15
    Different colors: 10/16 x 6/16 + 9/15 x 5/15
     
    Last edited: Nov 20, 2009
  2. jcsd
  3. Nov 20, 2009 #2

    LCKurtz

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    Looks good.
    How about subtracting out the ones where they sit together from the total?
    Circular permutations, same idea as 1.
    Think about combinations. How many ways can you choose 2 reds? 2 blues? So...
     
  4. Nov 21, 2009 #3
    Thanks LCKurtz, OK, here's what I got:

     
  5. Nov 21, 2009 #4

    LCKurtz

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    Remember the total number of ways 7 people can be arranged around a table is 6!, not 7!
    Hopefully you know that a fraction can't be the answer to how many ways to choose. I will ask you again: In how many ways can you pick 2 red marbles? 2 blue marbles? Think about combination formulas.
     
  6. Nov 22, 2009 #5

    So it's safe to say, 6! - (5! x 2) = 480

    Hmmmm
    2 red marbles = 2 x 10 = 20
    2 blue marbles = 2 x 6 = 12

    thanks
     
  7. Nov 22, 2009 #6

    LCKurtz

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    No. Have you studied permutations and combinations? Binomial coefficients? Do these symbols mean anything to you?

    [tex]C(n,r) =\binom{n}{r} = \frac{n!}{r!(n-r)!}[/tex]
     
  8. Nov 22, 2009 #7
    some permutations yes, but not that formula. However, I know a similar one.

    so back to the original problem, 10 red and 6 blue, in how many ways can you pick 2 of the same color?

    is it 16! / 10! 6! = 56

    I'm not sure how to compute picking different colors though....

    And are the others correct? Could you please give me the answer so I can fully grasp the concept?! This is driving me crazy and I don't want to fall behind with other stuff. I learn better by analyzing a correct problem, if that makes sense.

    Thanks!
     
  9. Nov 22, 2009 #8

    LCKurtz

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    Well, no, I won't just give you the answer. Surely the text you are using must have some discussion of how many ways you can choose r things from n things and what it has to do with the combination formulas.
     
  10. Nov 23, 2009 #9
    The book we use is VERY vague, with only a few examples, and we've only gone as far as the equation I've shown you. I don't understand why you cant help me with ONE answer, it's not like I haven't stressed myself trying to find the solution!! I mean you rather I never figure it out?
     
  11. Nov 23, 2009 #10

    LCKurtz

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