Suppose a lazy professor collects a quiz from each student in a class, then shuffles the papers and redistributes them randomly to the class for grading. How likely is it that no one receives his or her own quiz to grade?
How do I set it up?
I know this is all the possible permutation map to the number of students, but how do I set it up?
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$\begingroup$
combinatorics
discrete-mathematics
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1Relevant keyword: *derangement*. See e.g. [this](http://math.stackexchange.com/questions/399500/why-is-the-derangement-probability-so-close-to-frac1e) or [this](http://math.stackexchange.com/questions/2022673/permutation-without-fixed-points/2023219#2023219). – 2017-01-25
1 Answers
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We can get by inclusion exclusion that $!n=\sum\limits_{i=0}^n(-1) ^n \binom{n}{i}(n-i)!=n!\sum\limits_{i=0}^n \frac{(-1)^n}{i!}$.
So $\frac{!n}{n!}=\sum\limits_{i=0}^n \frac{(-1)^n}{i!}$. There is no closed formula for this, however if you look at the taylor series for $e^x$ it is clear that it approaches $e^{-1}$ when $n$ goes to infinity.