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Question about probability

  1. Nov 21, 2003 #1
    there are 10 questions. i can choose only exact 8 questions.

    what is the total combination?
     
  2. jcsd
  3. Nov 21, 2003 #2

    Njorl

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    45

    10!/[(8!)(2!)]

    Essentially, you have 10 choices for the first question you leave out, and 9 choices for the second question. This leads to 90 (10x9) combinations. But since it doesn't matter in which order you choose them, you've double counted (ie, you counted #1,#2 ans #2,#1 as different combinations), so divide by 2.

    Njorl
     
  4. Nov 21, 2003 #3

    jcsd

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    Njorl's given you the correct answer, but to expand this is a simple cominatrics question, your calculator will probably have a function that allows you to work it out, combinations are given by:

    [tex]_nC_r \equiv \left(\begin{array}{c}n\\r\end{array}\right) \equiv \frac{n!}{r!(n-r)!}[/tex]

    that is n different things taken r at a time, so in this case n = 10 and r = 8.
     
    Last edited: Nov 21, 2003
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