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Math Help - Probability of selecting 5 numbers from 1-40, no two being consecutive

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    Probability of selecting 5 numbers from 1-40, no two being consecutive

    Supposing we select 5 integers at random from 1 to 40 inclusive.

    How could we find the probability of none of these numbers being consecutive?

    I know we have 40 choose 5 outcomes total. I'm having trouble figuring out how many ways two or more of the 5 numbers would be consecutive though.
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  2. #2
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    Quote Originally Posted by fourierwarrior View Post
    Supposing we select 5 integers at random from 1 to 40 inclusive. How could we find the probability of none of these numbers being consecutive?
    I know we have 40 choose 5 outcomes total. I'm having trouble figuring out how many ways two or more of the 5 numbers would be consecutive though.
    How many ways to arrange 35 zeros and 5 ones so that no two ones are adjancent?
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  3. #3
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    Ok, so there's 40! ways of ordering the 5 1's and 35 0's total.

    I guess we could group the consecutive 1's as follows:

    5 1's [11111][000....00]

    35 ways

    4 1's [01111][010....00]

    5*35 ways??

    3 1's [11100][010...10]

    This way seems really difficult, I'm not really sure where to start.
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  4. #4
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    Here is a hint.
    How many ways to arrange 11100000 with no adjacent ones?
    _0_0_0_0_0_
    There are five zeros which create six places to put the ones.
    So the answer is _6C_3=\frac{6!}{3!\cdot 3!}.
    You appply that to this problem.
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  5. #5
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    Ohhhh! That makes sense.

    SO there are 35 zeros, which leaves 36 places to put the ones, so there are 36C5 ways to order the ones among the zeros.

    Hence, the probability I'm looking for is 36C5 / 40C5.

    Thank you!
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  6. #6
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    Quote Originally Posted by fourierwarrior View Post
    SO there are 35 zeros, which leaves 36 places to put the ones, so there are 36C5 ways to order the ones among the zeros.
    Hence, the probability I'm looking for is 36C5 / 40C5.
    You have it.
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