 Solution

Use the question's indexing convention: $m_{\lambda ,\mu }$ is the multiplicity of the irreducible Specht module $S_{\mathbb{C}}^{\mu }$ in the permutation module $M_{\mathbb{C}}^{\lambda }$. In notation for semistandard-tableau Kostka numbers this is often written $K_{\mu \lambda }$, with the indices reversed.

Let $S_{\lambda }=S_{{\lambda }_1}\times S_{{\lambda }_2}\times \cdots$ be the Young subgroup. The permutation character isand ${\xi }^{\lambda }(g)$ counts the $\lambda$-tabloids fixed by $g$. Equivalently it counts ordered partitions into labelled rows of the prescribed sizes, each row invariant under $g$. Character inner products and Frobenius reciprocity give the purely character-theoretic computationThe characters of symmetric groups are real-valued, so no conjugation is needed in the displayed subgroup average.

For $\lambda =(2,1^3)$, the Young subgroup has two elements, the identity and one transposition. Thus $m_{(2,1^3),\mu }=({\chi }_{\mu }(1)+{\chi }_{\mu }((12)))/2$, and part (a) givesThe resulting dimension is $1+3\cdot 4+3\cdot 5+3\cdot 6+2\cdot 5+4=60=5!/2$, the dimension of the permutation module.

This is the character average for Kostka numbers.