2

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?

Community
  • 1
Thomas W.
  • 133
  • 1
    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. – Brian M. Scott Dec 14 '12 at 08:07
  • @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?) – Thomas W. Dec 14 '12 at 08:51
  • Sure, I can do that: give me a couple of minutes. – Brian M. Scott Dec 14 '12 at 08:52

1 Answers1

1

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.

Community
  • 1
Brian M. Scott
  • 616,228