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The product principle

  1. Aug 29, 2007 #1
    1. The problem statement, all variables and given/known data
    In how many different ways can the top two positions be filled in a table tennis competition of 7 teams?
    How many 3-digit numbers can be formed using the digits 2, 3, 4, 5, and 6:
    a) as often as desired? b) once only?


    2. Relevant equations



    3. The attempt at a solution
    I can find the solution using the long way, but I'm wondering if there is a more logical and mathematical way to approach these equations?
     
  2. jcsd
  3. Aug 29, 2007 #2
    you know permutations and combinations?
     
  4. Aug 29, 2007 #3
    I got the first equation, and the b) section of equation 2, but how about a)? I'm confused by what they are asking.
     
  5. Aug 29, 2007 #4
    yes...that's what I used to solve for equation 1, and b), but a's wording is slightly confusing :p
     
  6. Aug 29, 2007 #5

    nrqed

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    You may use the same digit twice or even three times (like 665 or 666). That's all they mean.
    So how many choices do you have for the first digit? how many choices for the second? How many for the third? Multiply those three numbers and you are done.
     
  7. Aug 29, 2007 #6
    Oh Ok. So just to clarify things, each number could be used five times...therefore 5^3...and when each value is used once, 5x4x3....which suggests that in the first box, any value of the series could be used, in the second box, what's left over, etc...is my explanation correct? If you have a better one, feel free to share it ;p.
     
  8. Aug 30, 2007 #7

    HallsofIvy

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    Please be careful of your wording: each digit (better than "number" which is confusing) can be used three times. What you really mean to say, I think, is that for each digit in the 3 digit number, you can choose from any of the 5 given digits. Yes, you are correct that then there are 53= 125 possible numbers. If you are only allowed to use each of the given digits once, then there are 5 choices for the first digit, 4 choices for the second digit, 3 choices for the third digit and so 5(4)(3)= 60 possible numbers.
     
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