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How can I calculate the amount of deranged anagrams for specific word. For example, the word: Anona. I know that there are $\frac{5!}{2!2!1!} = 30$ anagrams. How can I calculate the amount of derangements: $$ oanan\ noaan\ nanao\ naaon $$ Is there any general formula in order to deal with more sophisticated cases?

Thanks!

Vinyl_cape_jawa
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Bob_Bobb
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    Yes, there is a formula. See my answer here. – Jair Taylor Sep 28 '19 at 16:46
  • @JairTaylor Excellent answer. I didn't know about this, and was just starting to write an answer explaining rook polynomials when your comment appeared, just in time! – saulspatz Sep 28 '19 at 16:55
  • @saulspatz Thanks! It is a gem of a formula. – Jair Taylor Sep 28 '19 at 19:32
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    For this particular problem, you could define $A_i$ to be the set of arrangements in which the $i$th position is a fixed point. You could calculate the number of derangements by subtracting $|A_1 \cup A_2 \cup A_3 \cup A_4 \cup A_5|$ from the $30$ possible anagrams of ANONA by using the Inclusion-Exclusion Priniciple. – N. F. Taussig Sep 29 '19 at 08:58
  • @N.F.Taussig what would the formula for calculating the union of arrangements be? – Jash Shah Oct 02 '20 at 14:49
  • Also can we solve this by saying that a set Ai can be a combination of two positions? For example in this case we say Ai can be positions 1 and 3? – Jash Shah Oct 02 '20 at 18:37
  • @JashShah $$|A_1 \cup A_2 \cup A_3 \cup A_4 \cup A_5| = \sum_{i = 1}^{5} |A_i| - \sum_{1 \leq i < j \leq 5} |A_1 \cap A_2| + \sum_{1 \leq i < j < k \leq 5} |A_i \cap A_j \cap A_k| - \sum_{1 \leq i < j < k < l \leq 5} |A_i \cap A_j \cap A_k \cap A_l| + |A_1 \cap A_2 \cap A_3 \cap A_4 \cap A_5|$$ There are five terms in the first and fourth summations and ten terms in the second and third summations. As for $A_1 \cap A_3$, this means the first and third positions are fixed points, so we have a word of the form Axoxa, where x represents one of the remaining letters. – N. F. Taussig Oct 02 '20 at 20:18

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