Interesting solutions: Ochominoes 
COLORS GROUPED Ochominoes has 24 tiles, with the colors assigned to tiles by how many little squares each dioct piece has attached to it. The colors divide into 0, 1, 2, 3, 4, 5 and 6 squares, giving three groups of 2 colors and 3 groups of 6. Separating and grouping of colors is a favorite theme. Here are a few color groupings. Finding such solutions is hard! Can you find one where the 8 holes are symmetrically arranged?
Here are a couple of very cool colorgrouped solutions found by Adam Criswell, our staunch squire helping out at Ye Olde Gamery at the Maryland Renaissance Festival. The second one is especially handsome, containing both sticks and enclosing one color completely:
COLORS SEPARATED
SYMMETRICAL PAIRS
Interestingly, by rearranging the pairings, it's possible to make 7 congruent symmetrical pairs! Another challenge to explore involves mixed pairs, where one piece by itself is symmetrical, and its partner is nonsymmetrical, yet the two together are symmetrical. Since there are 12 symmetrical and 12 nonsymmetrical tiles in an Ochominoes set, can 12 mixed pairs be formed? Well, no. An exhaustive search showed that one of the symmetrical tiles simply would not accept a nonsymmetrical partner. The answer, then, is that 10 pairs are peaceably formed "mixed couples", and the last two pairs contain two symmetrical pieces and two nonsymmetrical pieces, respectively. Here is one such solution:
Notice that the second row has two congruent pairs, and the bottom row has four congruent symmetries. The first pair in the first row is rotational. The two pairs on the right side are the nonmixed pairs, and by an unexpected coincidence are congruent as well. What is also remarkable is that the joint lines of all the mixed pairs are diagonal while their symmetries are vertical. Isn't that strange?
Problem solved! As of December 26, 2017, a nonexistence proof has come from George Sicherman and his trusty program. So a 23tile snake is the maximum possible, and it has millions of solutions, leaving some 24th piece languishing unpaired. Thanks, George, for clearing up this question. Some answers are just so... disappointing. Still, we're now going to hunt down all the pieces that could be the loner. Here's a start:

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