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The following puzzle is from the October 1961 issue of the Eureka journal (published by The Cambridge University Mathematical Society).

Arrange 15 balls, 3 of each of 5 colours, in the triangular array below, so that no two of the same colour lie in any line parallel to any side of the triangle.

    O
   O O
  O O O
 O O O O
O O O O O

I found the phrase ‘3 of each of 5 colours’ a bit hard to understand. My question is, can anyone suggest a easier to understand wording of this phrase? Any other suggestions to make the question easier to understand would be appreciated. I hope to eventually share this puzzle with students in grade 9 or 10.

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7 Answers 7

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I think 9th and 10th graders would understand "3 of each of 5 colours" perfectly well but if not then just choose 5 colors for them.

"Arrange 15 balls, 3 red, 3 yellow, 3 green, 3 blue, 3 purple, in the ..."

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What's wrong with "We have 15 balls in 5 colors, with 3 balls for each color"?

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    $\begingroup$ Clear and straightforward. This should be the accepted answer. $\endgroup$
    – iBug
    Jul 24 at 7:04
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You have 5 sets of three balls. Each set is a different colour....

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You could try rearranging the sentence, as in:

"You have 5 different colours of balls, and 3 identical balls in each colour, making 15 total."

It's a bit clunkier, but easier to understand in my opinion.

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"Arrange 15 balls, 3 of each of 5 colours, in the …" would be easier to understand if the parenthetical phrase were explicitly parenthetical:

Arrange 15 balls (3 of each of 5 colours), in the …

or:

Arrange 15 balls (5 colours, 3 balls of each colour), in the …

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A different approach, use numbers instead of colors... I would give them a drawing like the one you showed but with bigger circles. Instructions:

  • In each circle, place a number between 1 and 5.
  • Each number cannot be used more than 3 times. (even better, omit this. I don't think there is a solution if a number is used more than three times.)
  • On any line parallel to the sides of the triangle, we cannot have a repeated number.
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  • $\begingroup$ Your answer doesn't bring any solution to the question but instead presents another challenge, feel free to ask your own question instead of posting it as an answer! (Welcome to Puzzling!) $\endgroup$
    – Auribouros
    Jul 25 at 13:29
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    $\begingroup$ The question was not to solve the puzzle but to provide a different wording. I'm not sure it is a different challenge by using numbers instead of colors. To me it seems like the same puzzle worded differently which in my view could address the OP's request. $\endgroup$ Jul 25 at 13:47
  • $\begingroup$ You're free to post a puzzle as long as it's not a complete copy of this one, you're free to add some constraints to it, or make the whole puzzle bigger. But your answer was more befitting being a comment, or a new question with variations! $\endgroup$
    – Auribouros
    Jul 25 at 13:52
  • $\begingroup$ Got it, thanks! $\endgroup$ Jul 25 at 13:56
  • $\begingroup$ "...I don't think there is a solution if a number is used more than three times.)" Hadn't thought about that, but you're right. There's no way to put four balls of one color without having two be on the same line. Incidentally, I posted solutions at projectpluto.com/temp/triangle.txt using numbers rather than colors or colours. I did think of it as being "the same puzzle worded differently". $\endgroup$
    – Bill Gray
    Jul 25 at 15:03
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I know you weren't asking for an actual solution to this puzzle, but I couldn't resist:

enter image description here

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    $\begingroup$ But how many unique solutions are there? Jk $\endgroup$
    – JLee
    Jul 23 at 18:35
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    $\begingroup$ @JLee Can't prove it, but I think this may be unique, (not counting reflections/rotations/color swaps). I tried a few times, and kept settling on this same symmetrical solution. Might be able to prove it exhaustively with a computer if I took the time to throw together some code. $\endgroup$ Jul 23 at 18:44
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    $\begingroup$ @ralphmerridew That doesn't work - just at a glance, you've got 2 B's on the right side of the triangle, 2 C's on the left. and 2 D's on the 2nd row from the left $\endgroup$ Jul 23 at 19:04
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    $\begingroup$ @ralphmerridew Okay, that's valid, but yours are both actually the same solution, (though different from mine), just rotated and color-swapped. Rotate the first one 120 degrees clockwise and the second 120 counter-clockwise, so you have a vertical column of A's in one and E's in the other. Swap some colors around and you'll see they're the same. So that's 2 solutions counting mine. $\endgroup$ Jul 23 at 20:29
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    $\begingroup$ You nerd-sniped me here. The possible solutions, and the logic for determining them, are given at projectpluto.com/temp/triangle.txt . $\endgroup$
    – Bill Gray
    Jul 25 at 4:02

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