1. ## number of derangements

How is it possible that D8 > 8! ?
8! is the number of all possible permutations of 8 elements, and D8 is the number of permutations such that no element is in its position.
So {set of derangements} $\subset$ {set of permutations}.
How can there be more derangements than permutations of the same number of elements?

2. ## Re: number of derangements

It isn't possible. What values do you get for $D_8$ and 8! ?

According to Wikipedia, Derangement - Wikipedia, the free encyclopedia, 8!= 40320 while $D_8= 14833$.

3. ## Re: number of derangements

Well this is embarrassing...

Thank you.

4. ## Re: number of derangements

Originally Posted by Yuuki
How is it possible that D8 > 8! ?
8! is the number of all possible permutations of 8 elements, and D8 is the number of permutations such that no element is in its position.
So {set of derangements} $\subset$ {set of permutations}.
How can there be more derangements than permutations of the same number of elements?
Here is a bit more. Look at the webpage in reply #2.

It is common to use the approximation, $\mathcal{D}(N)\approx \frac{N!}{e}~.$

That shows the impossibility.