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Counting derangements

Lets call the derangement function f for clarity. At f(n), there are n hats and n people. Everyone can choose from n-1 hats. Person 1 takes hat i from n-1 choices. Person i still has n-1 hats to choose from and everyone else has n-2 has to choose from (they can't choose their own hat or i).

Now we need two cases for what person i does. Think of this as

  1. Person i takes hat 1
  2. Person i doesn't take hat 1 and we don't know what they'll take until later

In case 2, we know person i doesn't take hat 1, but nothing more. Previously we knew that person i had n-1 choices, now he has n-2, just like everyone else. This means we can calculate f(n-1) for this case. In case 1, person i is no longer forbidden to take hat 1. In essence, we know that person i and person 1 have swapped hats and no longer need to be matched, thus f(n-2).

Either of these cases is possible, so we have a recurrence that multiplies the (n-1) choices by the possibility of either happenning, f(n) = (n-1)(f(n-1) + f(n-2))

Categories : Algorithm

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