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Math Help - Counting Problem

  1. #1
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    Counting Problem

    I simply don't know how to start.

    In how many ways can a set of 10 positive integers less than 101 (100 integers) be selected such that the difference between any 2 numbers in the set is at least 2.

    I know that if the "difference between 2" wasn't there it would be C(100,10). But with the difference there how the heck would you set it up?

    thanks
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  2. #2
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    Quote Originally Posted by bfpri View Post
    In how many ways can a set of 10 positive integers less than 101 (100 integers) be selected such that the difference between any 2 numbers in the set is at least 2.
    The idea is simple. We can say include 13~\&~15 among the ten because |13-15|\ge 2 but not 15~\&~16.

    So how many ways can we pick ten of those numbers so no two are consecutive?
    How many ways can you arrange a string of ten 1ís and ninety 0ís so no two 1ís are next each other?
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  3. #3
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    Quote Originally Posted by Plato View Post
    The idea is simple. We can say include 13~\&~15 among the ten because |13-15|\ge 2 but not 15~\&~16.

    So how many ways can we pick ten of those numbers so no two are consecutive?
    How many ways can you arrange a string of ten 1’s and ninety 0’s so no two 1’s are next each other?
    Thanks for the reply.
    That would be 90!/10!*80! if you split it into 10 "10"s and 90 "1"s.
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  4. #4
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    Actually it would be \dbinom{91}{10}.
    The ninety zeros create ninety-one places to place the ones.
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  5. #5
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    Hmm, still don't quite understand why it's 91.
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  6. #6
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    Simply using, 1111000000, four ones and six zeros.
    How many ways to arrange that string so no two ones are together?
    Look at \_\_0\_ \_0\_ \_0\_ \_0\_ \_0\_ \_0\_\_ .
    Do you see there are seven places to put the four ones: \dbinom{7}{4}.
    So ninety zeros create ninety one places: \dbinom{91}{10}.
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  7. #7
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    Got it...thanks!
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