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Possible Duplicate:
I have a problem understanding the proof of Rencontres numbers (Derangements)

Given a vector of n elements, how can I calculate the number of "true" permutations, i.e. permutations that do not leave any element at the same position? Expressed differently, how do I find the number of possible $n\times n$ permutation matrices that only have zeros on their diagonal?

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marked as duplicate by Brian M. Scott, Davide Giraudo, user18119, martini, TMM Dec 14 '12 at 12:33

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    $\begingroup$ Such permutations without fixed points are called derangements; they’ve been discussed here many times, and the Wikipedia article to which I linked has information on counting them. (The closed formulas aren’t very nice, but the recurrences aren’t bad at all.) This earlier question has some very comprehensive answers. $\endgroup$ – Brian M. Scott Dec 14 '12 at 8:07
  • $\begingroup$ @BrianM.Scott: Thanks for the answer! I tried searching for "constant" or "invariant", but didn't find anything. So, derangements is the term... (Would you consider adding this as a "real" answer rather than a comment so that I can accept it?) $\endgroup$ – Thomas W. Dec 14 '12 at 8:51
  • $\begingroup$ Sure, I can do that: give me a couple of minutes. $\endgroup$ – Brian M. Scott Dec 14 '12 at 8:52
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Such permutations without fixed points are called derangements; they’ve been discussed here many times, and the Wikipedia article to which I linked has information on counting them. (The closed formulas aren’t very nice, but the recurrences aren’t bad at all.) This earlier question has some very comprehensive answers.

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