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Math Help - Geometery Formula for number of edges

  1. #1
    Member OnMyWayToBeAMathProffesor's Avatar
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    Question Geometery Formula for number of edges

    HELLO, I REALLY NEED HELP TO FIND THE GEOMETERY FORMULA FOR THIS EQUATION. I FOULD OUT IT HAS TO BE AN EXPONOTIONAL FORMULA. HERE ARE THE SPEFICS...


    I have a figure, if it has one vertex, it has 0 edges,so...
    Vertex-----------edge
    1-----------------0
    2-----------------1
    3-----------------3
    4-----------------6
    5----------------10
    6----------------15
    7----------------28(i think, im not sure)
    8----------------(do not know, could not get)


    please help to find a formula where i could find the number of edges with the number of vertexes and vice versa. I am toatally lost. this is an un directed graph. If possible please include graphs so i may undersatnd. also is a formaul and a function the same cause it says to find the function, thanxs!!!
    Last edited by OnMyWayToBeAMathProffesor; October 5th 2006 at 07:26 PM. Reason: vertew and edge chart was not good
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  2. #2
    MHF Contributor Quick's Avatar
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    This is funny, my class just went over this.

    Connect all the vertices. Tell me, if there are 5 vertices, how many line segments touch each vertex? the answer is 4.

    If there are 6 vertices, then the answer is 5.

    If there are 7, then the answer is 6.

    Do you understand what I did (I have not answered the problem yet, I've merely gone through the first, most important part)? If not, then I will go into more detail...
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  3. #3
    Member OnMyWayToBeAMathProffesor's Avatar
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    respond to repley

    i have understood somewat of wat u said, please explain the whole, the formula please, thanxs
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  4. #4
    MHF Contributor Quick's Avatar
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    Quote Originally Posted by OnMyWayToBeAMathProffesor View Post
    i have understood somewat of wat u said, please explain the whole, the formula please, thanxs
    Draw out three vertices and connect them.

    Notice that when there are 3 vertices there are 2 segments for each vertex.

    So why isn't the rule 2x3, or 6 segments?

    Because you are counting each segment twice, so the answer is really (3x2)/(2), or 3 segments.

    Now draw a diagram with 4 vertices, there are 3 segments from each vertex, but each segment got counted twice, so divide by 2 and you'll find there are in all (4x3)/(2) segments, or 6 segments.

    Continuing on you can say that a diagram with n vertices would have n-1 segments from each vertex, but each segment got counted twice, so divide by 2 and you'll find there are n(n-1)/2=(n^2-n)/2 line segments.

    test it:

    (5^2-5)/2=(25-5)/2=20/2=10 that checks out.

    (6^2-6)/2=(36-6)/2=30/2=15 that checks out.

    So the expression works!
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  5. #5
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    Q.How many ways are to draw edges?
    A.Since an edge is determined by 2 vertices then,
    nC2
    Which is,
    n!/2!(n-2)!=n(n-1)/2 which is what Quick has.

    This is the number of edges in a complete graph.
    It is not possible to determine the number edge in a general graph, since it can be anything.
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  6. #6
    Member OnMyWayToBeAMathProffesor's Avatar
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    replaey to previous post

    i got the part before the equal sign but wat about after the equal sign, how do u explain that?thanxs (this is for Quick)
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  7. #7
    Member OnMyWayToBeAMathProffesor's Avatar
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    For ThePerfectHacker

    wat are the ! for? thanxs
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  8. #8
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    Quote Originally Posted by OnMyWayToBeAMathProffesor View Post
    wat are the ! for? thanxs
    That is the factorial.

    n!=n(n-1)(n-2)...(2)(1)
    For example,
    4!=4*3*2*1=24

    The formula I mentioned.
    Is the number of combinations.
    Thus, I was looking for the number of combination involving 2 verticies.

    If the question was a digraph (which you purposely omit) then it can go both ways. Thus twice as much.
    n(n-1)
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  9. #9
    Member OnMyWayToBeAMathProffesor's Avatar
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    Talking Thanxs To Every One That Helped Me

    THANXS TO EVERY ONE THAT HELPED ME!!!!!!!!!!!!!!
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  10. #10
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    Quote Originally Posted by OnMyWayToBeAMathProffesor View Post
    THANXS TO EVERY ONE THAT HELPED ME!!!!!!!!!!!!!!
    You are warned for Caps lock abuse.
    You are warned for Smiley abuse.
    -=USER WARNED=-
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