Timeline for Non-enumerative proof that there are many derangements?
Current License: CC BY-SA 3.0
5 events
| when toggle format | what | by | license | comment | |
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| Jan 23, 2012 at 18:32 | comment | added | Timothy Chow | It doesn't even have anything to do with group theory per se. The probability that a random permutation fixes any particular element is $1/n$. So the expected number of fixed points is 1, by linearity of expectation. | |
| Jan 21, 2012 at 11:42 | comment | added | Marc van Leeuwen | Actually the average number of fixed points requires non computation at all: for any finite group acting on a set, the average number of fixed points is equal to the number of orbits (Burnside's lemma). | |
| Jan 20, 2012 at 13:47 | history | edited | Barry Cipra | CC BY-SA 3.0 |
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| Jan 19, 2012 at 21:05 | comment | added | Omer | The first three moments for the number of fixed points suffice. (The three first terms from inclusion-exclusion give a lower bound of $1/3$ for the fraction of derangements. | |
| Jan 19, 2012 at 18:07 | history | answered | Barry Cipra | CC BY-SA 3.0 |