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Math Help - Some more factorial word problems

  1. #1
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    Some more factorial word problems

    I find these a bit more complicated from the problem in my previous post. I can't seem to apply that logic to these problems.

    1. How many 3-digit positive integers are odd and do not contain the digit "5"?

    2. A talent contest has 8 contestants. Judges must award prizes for first, second, and third places. If there are no ties, in how many different ways can the 3 prizes be awarded?

    I'm sure these are so simple, but I've tried many different approaches and I keep getting wrong answers. (I'm studying for a standardized test...)

    Thank you.
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  2. #2
    Super Member Failure's Avatar
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    Quote Originally Posted by alaskamath View Post
    I find these a bit more complicated from the problem in my previous post. I can't seem to apply that logic to these problems.

    1. How many 3-digit positive integers are odd and do not contain the digit "5"?
    Again, imagine yourself selecting digits so as to get a number d_3 d_2 d_1 that satisfies the given requirements (where d_{1,2,3} are the digits).
    In order to make sure that the number will be odd, you must choose an odd digit for d_1. There are 5 odd digits, but since the digit 5 must not occur anywhere in the number you are about to construct you have only 5-1=4 possibilities for d_1.
    For d_2 there is no restriction placed on the digit, except that it must not be 5. So there are 10-1=9 possibilities for your choice of d_2.
    Now for digit d_3: again, it must not be 5 but also, since it is the leading digit of your number, you must not make it 0 either. So there are 10-1-1=8 possible choices for d_3.
    Since these choices are independent of each other, the total number of possibilities is the product 8\cdot 9\cdot 4

    2. A talent contest has 8 contestants. Judges must award prizes for first, second, and third places. If there are no ties, in how many different ways can the 3 prizes be awarded?
    Again, imagine yourself selecting the 3 highest placed persons by yourself. For the first place, you have 8 possibilities. For each of these 8 possibilities you have 8-1=7 possibilities to choose a person for the second place. And, finally, you have 8-2=6 possibilities to choose a person for the third place. This gives 8\cdot 7\cdot 6 different ways in all.
    Last edited by Failure; August 8th 2009 at 09:24 PM.
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