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Say I have 10 coins in a row, and I'm allowed to swap the positions of two coins at a time. Can I reverse the order of the coins by swapping each pair of positions once and only once (for 10 coins, 45 swaps)? I'm stumped, but I figure this is an already-thought-about area of math? Does this problem have a name I can search for?

Many thanks, Erik

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  • $\begingroup$ This isn't number theory, it falls under combinatorics or group theory. $\endgroup$ Mar 28, 2013 at 8:28

1 Answer 1

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You can:

$(1\;2)(1\;3)(1\;4)(1\;5)(1\;6)(1\;7)(1\;8)(1\;9)(1\;10)\cdot\\\ (2\;3)(2\;4)(2\;5)(2\;6)(2\;7)(2\;8)(2\;9)(2\;10)(3\;4)\cdot\\\ (3\;5)(3\;6)(3\;7)(3\;8)(3\;9)(3\;10)(4\;5)(4\;6)(4\;7)\cdot\\\ (4\;8)(4\;9)(4\;10)(5\;6)(5\;7)(5\;8)(5\;9)(5\;10)(6\;7)\cdot\\\ (6\;8)(6\;9)(6\;10)(7\;8)(7\;9)(7\;10)(8\;9)(8\;10)(9\;10)\\\ =\\\ (1\;2\;3\;4\;5\;6\;7\;8\;9\;10)(2\;3\;4\;5\;6\;7\;8\;9\;10)\cdot\\\ (3\;4\;5\;6\;7\;8\;9\;10)(4\;5\;6\;7\;8\;9\;10)(5\;6\;7\;8\;9\;10)\cdot\\\ (6\;7\;8\;9\;10)(7\;8\;9\;10)(8\;9\;10)(9\;10)\\\ =\\\ (1\;10)(2\;9)(3\;8)(4\;7)(5\;6)$

The construction obviously generalizes...

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