What is the total number of permutations of n digits where all digits will exchange their positions?

Asked Nov 13 '13 at 02:31

Active Nov 13 '13 at 02:31

Viewed 23 times

As we know, Total number of permutations of the digits 12345 is 5!.

Well, I am looking for another interesting fact. How many permutations among them are in the way that no digit is in its original position?

(For example, 54231 is one of such ones, where all the digits are in different positions from the original permutation 12345)

Thanks in advance. :-)

asked Nov 13 '13 at 02:31

2

search for the term "derangements" – Artem Nov 13 '13 at 02:34
This is just the number of derangements. As n grows, the probability approaches $1/e$, and it approaches that ratio very fast. – user99680 Nov 13 '13 at 02:34

0 Answers0