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Math Help - [Probability] Seating arrangement & Selection

  1. #1
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    [Probability] Seating arrangement & Selection

    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
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  2. #2
    Eater of Worlds
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    Your #1 is correct. For #2, subtract #1 from the total number of arrangements,which is 7!.

    For the circular. There are (n-1)! ways to arrange n people around a circular table. Use the same ideas you had on the first problem where they're in a line.
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  3. #3
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    Hello, CountNumberla!

    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 . . . . Right!

    Different colors: 10/16 x 6/16 + 9/15 x 5/15 . . . . no

    \text{Same colors: }\;\left(\frac{10}{16}\cdot\frac{9}{15}\right) + \left(\frac{6}{16}\cdot\frac{5}{15}\right) \quad=\quad \frac{1}{2}
    . . . . . . . . . . . {\color{red}RR} \quad\;\; \text{or} \qquad {\color{blue}BB}


    \text{Different colors: }\;\left(\frac{10}{16}\cdot\frac{6}{15}\right) + \left(\frac{6}{16}\cdot\frac{10}{15}\right) \quad=\quad \frac{1}{2}
    . . . . . . . . . . . . . {\color{red}R}{\color{blue}B} \quad\;\;\text{ or } \quad\;\; {\color{blue}B}{\color{red}R}

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  4. #4
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    Thanks everybody!

    Ok, here are my updated solutions:

    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:
    7! - (6! x 2) = 3600
    -------------------------------------------------------------------------------------------------------
    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
    7! - (5! x 2) = 4800
    -------------------------------------------------------------------------------------------------------
    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 = 1/2
    Different colors: 10/16 x 6/15 + 6/16 x 10/15 = 1/2
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