DOWNLOAD-DESKTOP-NOTEBOOK.nb (1.13 MB)
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L_1,...,L_{12} has 12! permutations, but, for the problem at hand, we need to think about how G_1, G_2, G_3, G_4 has 4! permutations and each group has 3! permutations within it. So, ultimately, we have 12! (4!)^{-1} (3!)^{-4} = 15400 functionally distinct arrangements of any twelve-letter set.