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Math Help - How many functions [5]--->[5] have at least one fixed point?

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    Member oldguynewstudent's Avatar
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    How many functions [5]--->[5] have at least one fixed point?

    If f is a function and f(i) = i then we call i a fixed point of f.

    a) How many functions [5] \rightarrow[5] have at least one fixed point?

    I have been working on this for hours! I've written out [3] \rightarrow[3], and [4] \rightarrow[4].

    I came up with this bizzare formula but I am sure it is incorrect. Please help!

    [\sum_{k=0}^{n-3}(n-k)(n-2)]+1
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by oldguynewstudent View Post
    If f is a function and f(i) = i then we call i a fixed point of f.

    a) How many functions [5] \rightarrow[5] have at least one fixed point?

    I have been working on this for hours! I've written out [3] \rightarrow[3], and [4] \rightarrow[4].

    I came up with this bizzare formula but I am sure it is incorrect. Please help!

    [\sum_{k=0}^{n-3}(n-k)(n-2)]+1
    How about thinking of it this way: what you want is the total number of functions minus the ones with NO fixed points.

    How far can you get with that conceptualization?
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  3. #3
    Member oldguynewstudent's Avatar
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    Quote Originally Posted by Jhevon View Post
    How about thinking of it this way: what you want is the total number of functions minus the ones with NO fixed points.

    How far can you get with that conceptualization?
    I have been working it from that angle also. I came out with 4^5 with no fixed points, but I could not justify that either.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by oldguynewstudent View Post
    I have been working it from that angle also. I came out with 4^5 with no fixed points, but I could not justify that either.
    yes, you have 5 elements, each of which has 4 choices of what to map to, that is, 4^5 functions. So the overall answer would be...?
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    Member oldguynewstudent's Avatar
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    Quote Originally Posted by Jhevon View Post
    yes, you have 5 elements, each of which has 4 choices of what to map to, that is, 4^5 functions. So the overall answer would be...?
    Yes thanks! That is the original answer I came up with 2 or 3 hours ago but then I thought it was wrong! 5^5 - 4^5
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by oldguynewstudent View Post
    Yes thanks! That is the original answer I came up with 2 or 3 hours ago but then I thought it was wrong! 5^5 - 4^5
    Yup, that's it
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