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Math Help - Probability

  1. #1
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    Probability

    Hi to all
    I have few short questions if you would like to look at them
    a)in how many ways can 3 boys and 3 girls sit in a row?
    b)in how many ways can 3 boys and 3 girls sit in the row if the boys and the girls are each to sit together
    c)in how many ways if only the boys must sit together?
    d)in how many ways if no two people of the same sex are allowed top sit together?
    Thank you
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  2. #2
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    What have you tried on any of these?
    Have you done any work on them?

    The answer to part c is: 2(3!)(3!).
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  3. #3
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    thanks
    Yes i done part a b and d.I just needed answere for c. I wanted just check the answeres.Thanks
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  4. #4
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    Actually i've got defferent answere for c which probably is wrong but i've got the same answere what you have for part b
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  5. #5
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    Actually parts b & c have the same numerical answer.
    The reasons for that are entirely different.
    You may be reading “if only” as “only if”. They are not the same!

    In the future, if you post your answers we will know that you are not just on a fishing expedition to see if someone will do your work for you.
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  6. #6
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    Hello, mauro21pl!

    a) In how many ways can 3 boys and 3 girls sit in a row?
    With six people, there are: . 6! = 720 ways.


    b) In how many ways can 3 boys and 3 girls sit in the row
    if the boys and the girls are each to sit together?
    There are two arrangements of the genders: . BBBGGG or GGGBBB

    In each, the three boys can be ordered in 3! ways
    . . and the girls can be ordered in 3! ways.

    Therefore, there are: . 2 \times 3! \times 3! \:=\;72 ways.



    c) In how many ways if only the boys must sit together?
    Duct-tape the boys together.
    Then there are four "people" to arrange. .There are 4! arrangements.

    For each arrangement, the three boys can be ordered in 3! ways.

    Therefore, there are: . 4! \times 3! \:=\:144 ways.



    d) In how many ways if no two people of the same sex are allowed to sit together?
    Since the boys and girls must alternate,
    . . there are two arrangements: . BGBGBG or GBGBGB

    For each arrangement, the three boys can be ordered in 3! ways
    . . and the three girls can be ordered in 3! ways.

    Therefore, there are: . 2 \times 3! \times 3! \:=\:72 ways.

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