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Math Help - 3-lists taken from [3]

  1. #1
    Member oldguynewstudent's Avatar
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    3-lists taken from [3]

    Consider the 3-lists taken from [3]. How many are there in which each element of [3] appears at least once?

    I get _{3}P_{3}.

    Answer the same question, but for 4-lists and 5-lists taken from [3].

    For the 4-list I get _{3}P_{3}*3.

    For the 5-list I get _{3}P_{3}* 3^2.

    Are these answers correct?
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  2. #2
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    Quote Originally Posted by oldguynewstudent View Post
    Consider the 3-lists taken from [3], 4-lists and 5-lists taken from [3].?
    Please define those terms. I have never seen them before.
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  3. #3
    Member oldguynewstudent's Avatar
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    Quote Originally Posted by Plato View Post
    Please define those terms. I have never seen them before.
    [3] is the set of all positive integers less than or equal to 3.

    The 3-lists would then be 123 132 213 231 312 and 321.

    A list is an ordered sequence of objects, and a k-list is a list of length k.
    Lists are also called words. Can also be written (1,2,3) etc. Without the constraint that the elements can only appear once, (1,1,1) would also be a 3-list of [3].
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    Quote Originally Posted by oldguynewstudent View Post
    [3] The 3-lists would then be 123 132 213 231 312 and 321.
    A list is an ordered sequence of objects, and a k-list is a list of length k. Lists are also called words. Can also be written (1,2,3) etc. Without the constraint that the elements can only appear once, (1,1,1) would also be a 3-list of [3].
    From that it seems to me that there are 3^3 3-lists from [3].
    And 3^4 in a 4-list from [3].
    Last edited by Plato; May 18th 2010 at 08:09 AM.
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    Member oldguynewstudent's Avatar
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    Quote Originally Posted by Plato View Post
    From that it seems to me that there are 3^3 3-lists from [3].
    And 3^4 in a 4-list from [3].
    Yes, that would be true except for the constraint that every element must appear at least once. ( I slightly mistated this constraint in my prior post.)

    Since every element must appear at least once, for the 3-list the lists would be (1,2,3) (1,3,2) (2,1,3) (2,3,1) (3,1,2) and (3,2,1); without the constraint we could add (1,1,1) (1,1,2) (1,1,3) etc.
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    Quote Originally Posted by oldguynewstudent View Post
    Yes, that would be true except for the constraint that every element must appear at least once. ( I slightly mistated this constraint in my prior post.)
    Well then, I think in the case of 4-list from [3] we want to count the number of surjections from a set of four to a set of three. There are 36 of those.
    Now with this understanding <3,2,1,1> is different from <1,2,3,1>.

    If it means that those are not different then the answer in \binom{1+3-1}{1}=3.
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