 Solution

ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2019/discrete-mathematics-continuous-mathematics-and-probability-resit/2/a/solution

There are

n

n - 1

choices for the second, and so on down to one choice. The multiplication principle therefore gives $n!$ permutations. This also agrees with the empty permutation convention

0! = 1

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