Ytukyg

given that $s_lambda s_mu=sum_{nu} C_{lambda mu}^nu s_nu$ with $vert lambdavert +vertmu vert=vert nuvert$, why does apparently also hold that $$h(lambda) h(mu) {vert nuvert choose vert lambda vert}=sum_{nu} C_{lambda mu}^nu h(nu)$$ 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?

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

Consider the relation
$$s_lambda s_mu=sum_nu C_{nu}^{lambdamu} s_nu.tag{1}$$
Take the coefficient of $x_1x_2cdots x_{r+t}$ in $(1)$.
From the right-hand side, this coefficient is
$$sum_nu C_{nu}^{lambdamu}h(nu).$$
But there are $binom{r+t}r$ pairs of monomials of degrees $r$ and $t$ that
multiply to give $x_1x_2cdots x_{r+t}$. As the Schur functions are symmetric
the coefficient from the LHS is $binom{r+t}r$ times that coefficient
of $x_1cdots x_r$ in $s_lambda$ times the coefficient of
of $x_1cdots x_t$ in $s_mu$, that is
$$binom{r+t}r h(lambda)h(mu).$$