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Math Help - permutation!

  1. #1
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    permutation!

    How many 3-digit odd numbers can be formed with the digits 1, 2, 3, . . . , 9, if no digit is repeated in any
    number?

    pls help on the problem above, thanks.
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  2. #2
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    Re: permutation!

    To get a number ending in 1- first can be any of 8 and second any of 7. So there are 8x7=56. But it will be the same for numbers ending in 3,5,7 or 9.
    Hence answer is 56x5= 280
    Thanks from lawochekel
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  3. #3
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    Re: permutation!

    Can you think of a sequence of steps to construct such a number? E.g., "at first I choose the last digit, then ..." Try to calculate in how many ways each step can be done. Then usually you have to multiply the numbers for each step. It is important that all possible numbers can be constructed using these steps, and no number is constructed twice using different choices.
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  4. #4
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    Re: permutation!

    Surely that is exactly what I have done.
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  5. #5
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    Re: permutation!

    Quote Originally Posted by biffboy View Post
    Surely that is exactly what I have done.
    Yes. The main reason I made a post is that I did not see your post. Also, I wanted to give a hint rather than a solution.
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  6. #6
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    Re: permutation!

    Quote Originally Posted by lawochekel View Post
    How many 3-digit odd numbers can be formed with the digits 1, 2, 3, . . . , 9, if no digit is repeated in any
    number?

    pls help on the problem above, thanks.
    N=5 \cdot \frac{8!}{(8-2)!}=5 \cdot 8 \cdot 7 =280
    Thanks from lawochekel
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