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Removing one coin from each of the triangles — more specifically, one a, one b, and one c — forms a pattern (like the example shown).

enter image description here

Every possibility is listed below (ignore the circles and squares around some of them).

enter image description here

How can you systematically list all of the possibilities, so that you get that answer?

Pier
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  • the figure should follow some kind of symmetry after the removal of some coins? – h-squared Mar 27 '20 at 10:54
  • I don't really understand what the space of all the possibilities is. Is the starting picture always the same? (The figure on the left) – Matti P. Mar 27 '20 at 10:57
  • This is not clear at all. Must you choose one coin from each of the three little triangles or could you take $a,b,c$ from a single little triangle? What are the rules? I expect that if you write the rules clearly the solution will emerge. – lulu Mar 27 '20 at 11:03
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    so you want to list the 27 possibilities? number the $a$ a1, a2, a3 and same for other letters, then take one from each – Gareth Ma Mar 27 '20 at 11:03
  • Assuming you are asking just for a way of picking one of the three a coins, one of the three b coins, and one of the three c coins with no further restrictions... simply list the numbers $000_3$ to $222_3$ in base $3$ and interpret the number as such a selection... the first digit corresponding to if you took the top copy (0), the bottom left copy (1), or the bottom right copy (2) of a respectively, and similarly so for the second digit and third digit... – JMoravitz Mar 27 '20 at 12:04
  • @h-squared no, the figure shouldn't follow any kind of simmetry – Pier Mar 27 '20 at 14:18
  • @lulu editing the question – Pier Mar 27 '20 at 14:20

1 Answers1

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We have to derive the underlying structure of the problem, forgetting about the artistic paraphernalia. We have three lines of $a$,$b$, $c$ and should select one element from each numbered line and one element from each letterized column: $$\left[\matrix{a_1&b_1&c_1\cr a_2&b_2&c_2\cr a_3&b_3&c_3\cr}\right]\ .$$ As in a $3\times3$-determinant there are $6$ ways to do this: Select one object from the first line ($3$ ways), then one allowed element from the second line ($2$ ways) and then the only remaining element from the third line.

You can arrange the admissible selection lexicographically as follows: $$a_1b_2c_3,\ a_1b_3c_2, \ b_1a_2c_3, \ b_1c_2a_3, \ c_1a_2b_3, \ c_1b_3a_2\ .$$

Christian Blatter
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