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Math Help - is this formula faulty.

  1. #1
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    is this formula faulty.

    the number of triangles that can be drawn four points of which no three are collinear is given by 4C3. but consider the image given. we can easily obtain that there are 8 triangles.could you please explain?
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  2. #2
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    Re: is this formula faulty.

    Hello, anigeo!

    The number of triangles that can be drawn connecting four points,
    of which no three are collinear, is given by: _4C_3

    But consider the image given.
    We can easily obtain that there are 8 triangles.
    Could you please explain?

    The key phrase is in blue.

    Using the four points, there are four triangles.

    If you use a fifth point or a sixth, there are, of course, more triangles.

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  3. #3
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    Re: is this formula faulty.

    but don't you see more than 4(8) triangles possible in the attached image.
    Attached Thumbnails Attached Thumbnails is this formula faulty.-untitled.png  
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  4. #4
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    Re: is this formula faulty.

    Soroban was referring to the use of non-collinear points (no 3 points on a straight line).
    In your image, you have 5 points.
    If no 3 were collinear, you'd have 5C3 = 10 triangles.
    However, in your diagram, 3 points are collinear along both diagonals,
    so you lose 2 of those 10.

    Unfortunately, where the diagonals cross has to be counted as a 5th point,
    since you are using it in the formation of half of those 8 triangles.
    (The formula counts triangles connected by points,
    not including the formation of new points where lines cross).
    Last edited by Archie Meade; October 22nd 2011 at 03:38 AM.
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  5. #5
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    Re: is this formula faulty.

    Quote Originally Posted by Archie Meade View Post
    Soroban was referring to the use of non-collinear points (no 3 points on a straight line).
    In your image, you have 5 points.
    If no 3 were collinear, you'd have 5C3 = 10 triangles.
    However, in your diagram, 3 points are collinear along both diagonals,
    so you lose 2 of those 10.

    Unfortunately, where the diagonals cross has to be counted as a 5th point,
    since you are using it in the formation of half of those 8 triangles.
    (The formula counts triangles connected by points,
    not including the formation of new points where lines cross).
    thanks i think i got it.it's not about the new points formed.
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