Find the sum of the permutations of degree $n$. For example, $n = 3$ would give: $123 + 132 + 213 + 231 + 321 + 312 = 1332$
I posed this problem to myself, and was able to solve it (I have put my work below). However, I did not at all utilize the symmetric group of degree $n$ nor (directly) any properties of group theory. I'm not sure that it's going to simplify things at all to do so, but I'm very rusty on group theory at this point and was hoping to see an alternate proof which perhaps uses some of those ideas.
My Solution:
There are $n!$ total permutations, and each number appears in a particular decimal place $(n-1)!$ times.
More concretely, The number $m$ will appear in the $j^{th}$ decimal place $(n-1)!$ times for $1 \le m, j \le n$
Every permutation of degree $n$ can be written as $$P_1 = \sum_{k = 0}^{n-1} a_k^{(1)} 10^k = a_0^{(1)} + 10 a_1^{(1)} + ... + 10^{n-1}a_{n-1}^{(1)}$$ Where all coefficients $a_j^{(1)}$ are distinct elements of the set $\{1,2,...,n\}$
Take another distinct permutation, $$P_2 = \sum_{k = 0}^{n-1} a_k^{(2)} 10^k = a_0^{(2)} + 10 a_1^{(2)} + ... + 10^{n-1}a_{n-1}^{(2)}$$ Adding them: $$P_1 + P_2 = (a_0^{(1)} + a_0^{(2)}) + 10(a_1^{(1)} + a_1^{(2)}) +...+ 10^{n-1}(a_{n-1}^{(1)} + a_{n-1}^{(2)})$$ From this, it is clear that summing over all $n!$ permutations yields: $$P = \sum_{k = 1}^{n!} a_0^{(k)} + 10 \sum_{k=1}^{n!} a_1^{(k)} +...+ 10^{n-1} \sum_{k=1}^{n!} a_{n-1}^{(k)} = \sum_{j=0}^{n-1} 10^j \sum_{k=1}^{n!} a_j^{(k)}$$ Since each digit appears in the $j^{th}$ decimal place $(n-1)!$ times, each sum $\sum_{k=1}^{n!} a_j^{(k)}$ has the same value (it sums the same numbers in a different order), so that it must be independent of $j$. Furthermore, this value is easily determined: $$\sum_{k=1}^{n!} a_j^{(k)} = (n-1)! + 2(n-1)! + ... + n (n-1)! = (n-1)! \sum_{k=1}^n k = (n-1)!\frac{n(n+1)}{2} = \frac{(n+1)!}{2}$$
Therefore, $$P = \frac{(n+1)!}{2}\sum_{j=0}^{n-1} 10^j = \frac{(n+1)!}{2} \frac{10^n -1}{9} = \frac{(n+1)!(10^n - 1)}{18}$$