Tessellate S.T.E.M.S (2019) - Mathematics - Category B - Set 3 - Objective Problem 4

The average is $\frac1{10!} \sum_{\sigma \in S_{10}} (|\sigma(1) - \sigma(2)| + \cdots + |\sigma(9)-\sigma(10)|) = \frac1{10!} \sum_{\sigma \in S_{10}} |\sigma(1) - \sigma(2)| + \cdots + \frac1{10!} \sum_{\sigma \in S_{10}} |\sigma(9) - \sigma(10)| = \frac5{10!} \sum_{\sigma \in S_{10}} |\sigma(1)-\sigma(2)|.$

Let $f(n)$ be the number of permutations in $S_{10}$ for which $|\sigma(1)-\sigma(2)| = n.$ Enumerating the possibilities shows that $f(9) = 2 \cdot 8!, f(8) = 4 \cdot 8!, \ldots, f(1) = 18 \cdot 8!.$ (For instance, $f(9)$ is the number of permutations with $\sigma(1) = 1, \sigma(2) = 10,$ which is $8!$ , plus the number of permutations with $\sigma(1) = 10, \sigma(2) = 1,$ which is another $8!.$ ) So $f(n) = 2(10-n)8!.$

Then the sum becomes $\frac5{10!} \sum_{n=1}^9 nf(n) = \frac5{10!} \sum_{n=1}^9 2n(10-n)8! = \frac19 \sum_{n=1}^9 n(10-n) = \frac{165}9 = \frac{55}3,$ and the answer is $\fbox{58}.$