1.3.12 problem 41

We can treat this problem as sampling numbers $1$ to $n$ with replacement with each number being equally likely. There are $n^n$ possible sequences. To count the number of sequences with exactly one of the numbers missing, we first select the missing number. There are $n$ ways to do this. The rest of the numbers have to be sampled at least once with one number being sampled exactly twice. There are $n-1$ choice to select the number that will be sampled twice. Finally, we have $n$ sampled numbers which can be ordered in any of $\frac{n!}{2}$ ways, since one of the numbers is repeated. Thus, the answer is