LR-rule and Standard Young Tableau counting

Question

given that $s_\lambda s_\mu=\sum_{\nu} C_{\lambda \mu}^\nu s_\nu$ with $\vert \lambda\vert +\vert\mu \vert=\vert \nu\vert$, why does apparently also hold that $$h(\lambda) h(\mu) {\vert \nu\vert \choose \vert \lambda \vert}=\sum_{\nu} C_{\lambda \mu}^\nu h(\nu)$$ where $h(\lambda)$ denotes the count of Standard Young Tableaux as given by the hooklength formula? A similar relation, but without the binomial factor, is well known for the semi-standard tableaux. This last one is easily understood by comparing monomials in the schur functions on both sides together with the relation between the weights of the monomials and semi-standard tableaux. For the relation above, I fail to grasp the link.
Any hints?

Answer 1

Write $r=|\lambda|$, $t=|\mu|$. Then $|\nu|=r+t$ for all the $\nu$ in the sum.

Consider the relation $$s_\lambda s_\mu=\sum_\nu C_{\nu}^{\lambda\mu} s_\nu.\tag{1}$$ Take the coefficient of $x_1x_2\cdots x_{r+t}$ in $(1)$. From the right-hand side, this coefficient is $$\sum_\nu C_{\nu}^{\lambda\mu}h(\nu).$$ But there are $\binom{r+t}r$ pairs of monomials of degrees $r$ and $t$ that multiply to give $x_1x_2\cdots x_{r+t}$. As the Schur functions are symmetric the coefficient from the LHS is $\binom{r+t}r$ times that coefficient of $x_1\cdots x_r$ in $s_\lambda$ times the coefficient of of $x_1\cdots x_t$ in $s_\mu$, that is $$\binom{r+t}r h(\lambda)h(\mu).$$