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 \b[$n!$ <permutations>]. This also agrees with the empty <permutation> convention $0!=1$.

Solution

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

 Solution (source code)