Re: number of derangements

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

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

Re: number of derangements

Well this is embarrassing...

I had somehow read the value on my calculator as 4320.

Thank you.

Re: number of derangements

Quote:

Originally Posted by

**Yuuki** How is it possible that D_{8} > 8! ?

8! is the number of all possible permutations of 8 elements, and D_{8} is the number of permutations such that no element is in its position.

So {set of derangements} $\displaystyle \subset$ {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, $\displaystyle \mathcal{D}(N)\approx \frac{N!}{e}~.$

That shows the impossibility.