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Counting question

  1. Sep 6, 2010 #1
    1. The problem statement, all variables and given/known data
    How many different bit strings can be formed using six 1s and eight 0s?



    2. Relevant equations

    C(n,r) = n!/(r!*(n-r)!)



    3. The attempt at a solution

    since theres six 1s and eight 0s there are 14 slots. So I'm guessing for the possible six 1s it would be C(14,6) and the remaining eight 0s would have to be C(8,8) since there are only 8 slots left after six 1s are chosen so it would be C(14,6)*C(8,8)?

    The last part C(8,8) seems kind of werid so I'm not very convinced this is correct. Any help is apperciated thanks.
     
  2. jcsd
  3. Sep 6, 2010 #2

    vela

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    Yeah, that's right. The factor of C(8,8), which equals 1, is correct, but usually you just gloss over that since you know the rest of the slots have to be filled with what's left.
     
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