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Math Help - Permutations

  1. #1
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    Permutations

    How many four-digit numbers can you form with the digits 1, 2, 3, 4, 5, 6, and 7 if no digit is repeated?

    How many of these four-digit numbers are odd numbers?

    How many of them are even numbers?


    I need help with the bolded questions. please explain, I have the answers from the textbook but I dont know how they got them.
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  2. #2
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    A. You have 7 items, and are meant to chose 4 of them to form a number. This is a permutation yes? How would you solve a permutation where you have N items and need to choose K?

    B. This is a "tricky" one that requires a quick bit of reasoning. For a number among the amount you found in part A to be odd, it should have the LAST digit as the 1, 3, 5 or 7. Correct? Let one of those numbers be our base. We are then left with 6 numbers and we need to choose 3 to pair with our odd number. Do you see where I am going with this?

    C. The same process as B.

    Let me know what you get.
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  3. #3
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    Quote Originally Posted by john-1 View Post
    1) How many four-digit numbers can you form with the digits 1, 2, 3, 4, 5, 6, and 7 if no digit is repeated?

    2) How many of these four-digit numbers are odd numbers?

    3) How many of them are even numbers?
    .
    #1 (7)(6)(5)(4) WHY? You explain that!

    #2 (4)(6)(5)(4) WHY? You explain that!

    #3 You do this one for yourself!
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  4. #4
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    b) 6P3 = 120 x 4 = 480

    c) 6P3 =120 x 3 = 360

    How's this look ANDS! ?



    Plato:
    how did you get that arrangement? 4 x 6 x 5 x 4?
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  5. #5
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    That's it. Just remember a lot of this stuff LOOKS more difficult than it really is. The toughest part of that problem is having that quick lightbulb go off and think "Oh heck, I know what an odd number looks like!"

    how did you get that arrangement? 4 x 6 x 5 x 4?
    (6)(5)(4) is the same is 6P3. The leading four is the fact that you have four odd-bases to choose from.
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