# number of derangements

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• May 28th 2013, 01:51 PM
Yuuki
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} $\displaystyle \subset$ {set of permutations}.
How can there be more derangements than permutations of the same number of elements?
• May 28th 2013, 02:27 PM
HallsofIvy
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 $\displaystyle D_8= 14833$.
• May 28th 2013, 02:42 PM
Yuuki
Re: number of derangements
Well this is embarrassing...
I had somehow read the value on my calculator as 4320.

Thank you.
• May 28th 2013, 02:45 PM
Plato
Re: number of derangements
Quote:

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} $\displaystyle \subset$ {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.