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 $n$ choices for the first position, then $n-1$ 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 $0!=1$.