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Math Help - Basic Question

  1. #1
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    Basic Question

    I need to know if I'm doing this problem accordingly

    A drawer contains n=5 different and distinguishable pairs of socks (total of 10). If a person randomly selects four socks what is the probability that there is no matching pair in the sample?

    solution:

    1-[(4)(\frac{4!}{10!}) - (4)(\frac{1}{10!}) + \frac{1}{10!}]
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  2. #2
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    Hello, lllll

    Using 10!, you seem to be working with permutations.
    . . Sorry, I don't understand your answer.


    A drawer contains 5 different and distinguishable pairs of socks (total of 10).
    If a person randomly selects four socks, what is the probability that
    there is no matching pair in the sample?
    There are: . {10\choose4} \:=\:{\color{blue}210} possible samples.


    Suppose the pairs of socks are of five different colors.
    To have no matching pairs, we must have four socks of different colors.

    There are: . {5\choose4} \:=\:5 choices for the four colors.
    Then we must select one sock from each of those pairs.
    . . There are: . {2\choose1}{2\choose1}{2\choose1}{2\choose1} \:=\:16 ways.

    Hence, there are: . 5\cdot16 \:=\:{\color{blue}80} ways to get four socks of different colors.


    Therefore: . P(\text{no matching pairs}) \;=\;\frac{80}{21} \;=\;\boxed{\frac{8}{21}}

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  3. #3
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    why would it not be a permutation, doesn't the order matter?
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