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Counting Permutations

  1. Sep 2, 2007 #1
    1. The problem statement, all variables and given/known data
    How many 3 digit numbers can be constructed from digits 1, 2, 3, 4, 5, 6, and 7 if each digit may be used once only and the number is odd?

    2. The attempt at a solution
    What number do they speak of? The resulting 3 digit number? How do I approach this equation?
  2. jcsd
  3. Sep 2, 2007 #2


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    For example, I could pick out the numbers 1,2 and 3 and form the number


    but also, I could form, 132 ,or 213, or 231, or 312, or 321.

    But, that is just one way to pick three numbers (1,2,3). I could have chosen to pick out the numbers 3,5 and 1. And I could then form 6 different numbers (135,153,315,351,513,531) with those.

    If I were you I would start off by thinking about how many different ways there are to choose three different things out of an array of 7 different things. "Seven choose three".

    Then, you know that for any set of three, you can make 6 numbers, but you have to figure out how many of them are odd. Good luck.
  4. Sep 2, 2007 #3


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    Oh. Yeah. They are probably talking about the *resulting* number (the three digit number). That is a very confusing way to word the problem. It is certainly vague.
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