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Math Help - pairs of shoes problem

  1. #1
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    pairs of shoes problem

    10 pairs of shoes(ie 20 shoes) pick 8.
    P(no complete pair)
    P(exactly 1 complete pair)



    P(exactly 1 complete pair)= 10C1(pick the pair) *18C1*16C1*14C1*12C1*.../20C8

    i dont know why this doesnt work? im picking a pair then picking each other shoe such that i dont pick its pair...?


    P(no complete pair)= 1-P(at least 1 complete pair)

    which i would do in the same fashion as above but its not working so im missing something.
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  2. #2
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    Quote Originally Posted by stumped765 View Post
    10 pairs of shoes(ie 20 shoes) pick 8.
    P(no complete pair)
    P(exactly 1 complete pair)
    Could you please explain the problem statement more. I can think of three interpretations

    1) We pick 8 shoes
    2) We pick 16 shoes, in pairs
    3) We pick 16 shoes, one by one

    The difference between (2) and (3) is:

    if we select 12 12 34 34 56 56 78 78

    then for (2) we do not have a complete pair; for (3) we have 8 complete pairs.
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  3. #3
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    were selecting 8 shoes from the 20shoes
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  4. #4
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    Quote Originally Posted by stumped765 View Post
    were selecting 8 shoes from the 20shoes
    Thanks.

    Quote Originally Posted by stumped765 View Post
    10 pairs of shoes(ie 20 shoes) pick 8.
    P(no complete pair)
    P(exactly 1 complete pair)



    P(exactly 1 complete pair)= 10C1(pick the pair) *18C1*16C1*14C1*12C1*.../20C8

    i dont know why this doesnt work? im picking a pair then picking each other shoe such that i dont pick its pair...?
    Shouldn't it be 10C1 * 9C6 * 2^6 / 20C8 ?

    We pick the complete pair (ten choices), then from the remaining nine pairs we must select six shoes, and no two shoes can be from the same pair, thus we are selecting six pairs (9C6) and beyond that we have two choices per pair, giving the 2^6 factor.

    Quote Originally Posted by stumped765 View Post
    P(no complete pair)= 1-P(at least 1 complete pair)

    which i would do in the same fashion as above but its not working so im missing something.
    I would think P(no complete pair) would be simpler to compute directly than P(at least 1 complete pair), and the computation would proceed similarly to how I described above.
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