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    OlympiadHardCombinatorics10–12

    Derangements

    Desarranjos

    How many permutations of {1,2,3,4,5}\{1, 2, 3, 4, 5\} have no element in its original position (derangements)?

    Use the formula D(n)=n!×∑(−1)kk!D(n) = n! \times \sum \frac{(-1)^{k}}{k!} for k=0k=0 to nn.

    Solution

    Step 1 of 5

    1. 1.D(5)=5!×(1−1+12−16+124−1120)D(5) = 5! \times \left(1 - 1 + \tfrac{1}{2} - \tfrac{1}{6} + \tfrac{1}{24} - \tfrac{1}{120}\right)