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Let \(d_n\) be the number of derangements of \(n\) objects. For example, using the techniques of this section, we find

\begin{equation*} d_3 = 3!-\left({3 \choose 1}2! - {3 \choose 2}1! + {3 \choose 3}0! \right)\text{.} \end{equation*}

We can use the formula for \({n \choose k}\) to write this all in terms of factorials. After simplifying, for \(d_3\) we would get

\begin{equation*} d_3 = 3!\left(1 - \frac{1}{1} + \frac{1}{2} - \frac{1}{6} \right)\text{.} \end{equation*}

Generalize this to find a nicer formula for \(d_n\text{.}\) Bonus: For large \(n\text{,}\) approximately what fraction of all permutations are derangements? Use your knowledge of Taylor series from calculus.

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