 Solution

The stabilizer of an ordered pair is the subgroup ${\Sigma }_{n-2}$ fixing both entries, so the permutation module is ${\mathrm{Ind}}_{{\Sigma }_{n-2}}^{{\Sigma }_n}1$. Induce in two stages through ${\Sigma }_{n-1}$. The first stage is $S^{(n-1)}\oplus S^{(n-2,1)}$. The characteristic-zero induction branching rule for a symmetric group adds one box in every possible partition position, with multiplicity one. Thus the two second-stage decompositions areBy transitivity of induction this yieldsThe dimensions are $1$, $2(n-1)$, $n(n-3)/2$ and $(n-1)(n-2)/2$, whose sum is $n(n-1)$, checking the decomposition also at $n=4$.