how would one go about doing this? I know the character tables of both groups and that the multiplicity of the irreducible representation $V$ in $W$ is
$$\frac{1}{|G|}\sum_{a\in G}\chi_V(a)^*\chi_W(a) $$ My initial thought was to divide the elements of $S_3$ into the various conjugacy classes of $S_4$, for example $(12)(3)$ is in the conjugacy class of $(12)(3)(4)$. The following classes of $S_4$ contain elements of $S_3$
$[(1)(2)(3)(4)]$
$[(123)(4)]$
- $[(12)(3)(4)]$
How should I proceed from here? I know in any case that all the elements in the conjugacy classes of $S_4$ that I listed above belong to some $S_3\subset S_4$, so I tried to use the formula by counting for $V$ an irreducible representation of $S_4$
$$\chi_V^{S_4}(a)= \chi_V^{S_3}(a)$$ if $a\in [(1)(2)(3)(4)],[(12)(3)(4)]\textrm{ or }[(123)(4)]$ and where by $\chi_V^{S_3}(a)$ I mean the character of the corresponding conjugacy class in $S_3$ (e.g. $[(12)(3)]$ for $[(12)(3)(4)]$), but this gives wrong results, for example asking how many times the trivial representation of $S_3$ appears in the trivial representation of $S_4$ yields
$$\frac{1}{24}(1+8+6)=\frac{15}{24} $$ because there is one element in the class $[(1)(2)(3)(4)]$, $6$ elements in the class $[(12)(3)(4)]$ and $8$ in the class $[(123)(4)]$, but this clearly doesn't make sense. I think I'm misunderstanding the question, what does it even mean to decompose a representation of a group into representations of another group?