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Thread: an intriguing puzzle

  1. #1
    Senior Member drunk's Avatar
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    an intriguing puzzle

    .......................................







    how many triangles in this image?
    I'll let people try and figure it out and won't post the spoiler until afterwards
    and just to be honest about it; I couldn't figure it out without the spoiler
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    Last edited by drunk; 04-05-2020 at 12:49 PM.
    but I don't care too much for money - money can't buy me love ..............................the Beatles

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    10

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    Random number herder Norm's Avatar
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    I counted 15 and then fell off my chair,
    "Croyez ceux qui cherchent la vérité, doutez de ceux qui la trouvent." --André Gide

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    Looks like more than 14, maybe 20.
    https://twitter.com/carnegiemellon/s...21728914153472

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    Random number herder Norm's Avatar
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    Yeah, but two of the lines do not intersect all of the other lines. That should mean it must be less than c(6,3)

    Never mind. They don't in the CMU thread, but they do here.
    Last edited by Norm; 04-06-2020 at 01:40 PM.
    "Croyez ceux qui cherchent la vérité, doutez de ceux qui la trouvent." --André Gide

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    Norm’s answer is correct. Since none of the lines are parallel, each selection of 3 lines determines a unique triangle. Thus
    there are C(6,3) triangles.

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    Random number herder Norm's Avatar
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    I'm not certain I'm correct. Too lazy to prove myself wrong.
    "Croyez ceux qui cherchent la vérité, doutez de ceux qui la trouvent." --André Gide

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    Senior Member drunk's Avatar
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    ........................




    the puzzle was created by the mathematician Po Shen-Loh



    the answer is 20



    from the article:



    "To solve the problem, start with this observation: Any three lines in the diagram define one and only one triangle. It follows that the total number of triangles will be equal to the number of three-line combinations that can be chosen from a group of six lines.

    How do you calculate that? Choose a line, any line. Since there are six lines, there are six choices. Next, choose a line to be the second side of the triangle. At first, you might think there are again six choices,but one line has already
    been chosen as a line for the first side, so only five choices are left. Likewise, for the third side of the triangle, there are four lines left to choose from.
    Thus, the total number of ways you can choose the sides of the triangle equals 6×5×4, or 120. Clearly, there are not 120 triangles in the diagram. That’s because all of those combinations are being counted more than once.
    For clarity, number the lines from 1 to 6, and look at the triangle defined by lines 1, 2 and 3. It’s the same triangle whether you choose line 1, then line 2, then line 3; or line 1, then line 3, then line 2.



    Indeed, there are as many ways to create that triangle as there are ways to choose lines 1, 2 and 3. Jumble the digits every which way: 123, 132, 213, 231, 312, 321. There are six possibilities; likewise, any triangle in the diagram could be created six possible ways.
    So now, divide out the redundancy. The total number of triangles created by those six lines is (6×5×4)/6, or 20. That’s the answer.
    Here’s where math becomes powerful. The same procedure works for any number of lines. How many triangles are created by seven nonparallel lines? That’s (7×6×5)/6, or 35. What about 23 lines? (23×22×21)/6, or 1,771. How about 2,300 lines? That’s (2300×2299×2298)/6, which is a big number: 2,025,189,100.
    The same calculation applies no matter how many lines there are. Compare that approach to brute-force counting, which is not only laborious and error-prone but provides no way to check the answer. Math produces the solution and the rationale for it.
    It also reveals that other problems are, at heart, identical. Put balls of six different colors into a bag. Pull out three. How many different possible color combinations are there? 20, of course.
    That’s combinatorics, and it’s useful for solving problems of this type. It comes with its own notation, to simplify the process of calculating, and involves a lot of exclamation points. The expression n! — “n factorial,” when said aloud — describes the product of multiplying all the integers from 1 to n. So 1! equals 1; 2! equals 2×1, or 2; 3! equals 3×2×1, or 6. And so on.
    In the problem by Dr. Loh, the calculation for the number of triangles can be rewritten like this: 6!/(3!3!)."





    https://www.nytimes.com/2019/08/21/s...es-pemdas.html



    Last edited by drunk; 04-07-2020 at 03:50 AM.
    but I don't care too much for money - money can't buy me love ..............................the Beatles

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    Senior Member drunk's Avatar
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    ..........................................








    okay, I admit, I don't completely understand this explanation

    here is my question:

    in the last paragraph of the explanation this is written:


    " Put balls of six different colors into a bag. Pull out three. How many different possible color combinations are there? 20, of course."


    how about if you pulled out just one ball so there were 5 left - how would you figure that out?

    how about if you pulled out just two balls so there were 4 left - how would you figure that out?

    help me out - I don't really get it





    I thought a little bit more and maybe I got it - if you pulled out just two balls so there were 4 left you get all the possible combinations by doing this:

    6 *5 *4 * 3 = 360.......... now divide 360 by 6 and get 60

    is that right? I think so but I'm not sure
    Last edited by drunk; 04-07-2020 at 07:45 AM.
    but I don't care too much for money - money can't buy me love ..............................the Beatles

  10. #10


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    In the original problem, he's just explaining that the combination of six things taken three at a time is (6x5x4)/(3x2x1) = 20.

    In your two problems, the answers are 60 each time.

    Don

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