Permutations of a set that keep at least one integer fixed

combinatorics permutations

Solution 1:

This is equal to $n!$ minus the number of permutations which keep no elements fixed. These are known as derangements and you can find all the relevant formulas at the Wikipedia article. You can count them, for example, using inclusion-exclusion.

If you just want the number of derangements of an $n$-element set, then a nice way to compute it is to round $\frac{n!}{e}$ to the nearest integer. For example, when $n = 5$ we want to round $\frac{120}{e} \approx 44.1$, so there are $44$ derangements, hence $120 - 44 = 76$ permutations which fix at least one element.

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