I'm trying to apply Mackey's theorem to a toy example where I start with the trivial $S_3\times S_2$-module $V$, induce to $S_5$ and restrict back to $S_3\times S_2$. The version of Mackey's theorem I'm using states $(V^G)_H = \bigoplus_{s \in H \setminus G/H} [s(V)_{H\cap sHs^{-1}}]^G$. Here $s(V)$ is notation for a module with vector space $V$ and action $g(s(V)) = sgs^{-1}(V)$. In my case $V$ is trivial so $s(V) = V$. Also $V^G$ denotes induction to $G$ and $V_H$ denotes restriction to $H$.
I believe one can count double cosets of the symmetric group by parabolics $S_\lambda,S_\mu$ by counting tableaux with shape $\lambda$ and weight $\mu$ (or viceversa). I find two tableaux and from them I get that two representatives for double cosets of $S_3\times S_2 \setminus S_5 /S_3 \times S_2$ are $1$ and $s_3 = (3, 4)$.
Now I calculate $({S_3\times S_2}) \cap s_3({S_3\times S_2})s_3 = \{1, s_1\} = S_2$.
Now according to Mackey's theorem I find that $(V^{S_5})_{S_3\times S_2}$ equals $(V_{S_3\times S_2})^{S_3\times S_2} \oplus (V_{S_2})^{S_3\times S_2} = V \oplus (V_{S_2})^{S_3\times S_2}$. But the dimension of the former is $[S_5:S_3\times S_2] = 10$ while the dimension of the latter is $1 + [S_3\times S_2:S_2] = 7$, so clearly something went wrong along the way.
I would like to know the mistake I'm making.
You have missed a double coset.
If we first work inside $S_4$, then we see that there are two double cosets $$ S_4 = S_3 \cup T \quad\textrm{where }T=S_3(34)S_3 $$ and $|T|=24-6=18$.
Thus in $S_5$ we have $$ H(34)H = T\cup T(45)\cup (45)T\cup (45)T(45), $$ so that $|H(34)H|\leq 4\cdot18=72$.