This is maybe a little basic for MathOverflow, but I'm hoping it will get some interesting answers.

Let $\unrhd$ be the dominance order on partitions of $n \in \mathbb{N}$.
For partitions $\lambda$ and $\mu$ of $n$, the *Kostka Number* $K_{\lambda\mu}$ is the number of semistandard Young tableaux of shape $\lambda$ and content $\mu$.

If $t$ is such a tableau then the $\mu_1+\cdots +\mu_r$ entries $k$ of $t$ such that $k \le r$ all lie in the first $r$ rows of $t$. Hence $\lambda_1 + \cdots + \lambda_r \ge \mu_1 + \cdots + \mu_r$ for each $r$ and so $\lambda \unrhd \mu$.

Is there a short combinatorial proof of the converse: if $\lambda \unrhd \mu$ then $K_{\lambda\mu} > 0$?

A constructive proof, maybe using the characterization of neighbours in the dominance order by single box moves on Young diagrams, would be especially welcome.

**Edit.** Using some representation theory it's possible to make this strategy work. The following is the symmetric functions version of Theorem 2.2.20 in the textbook by James and Kerber. Obviously $K_{\lambda\lambda} = 1$. Let $\mu$ and $\mu^\star$ be neighbours
in the dominance order, with $\mu \rhd \mu^\star$, so $\mu^\star_i=\mu_i-1$, $\mu^\star_j = \mu_j+1$ for some $i < j$, and $\mu^\star_k = \mu_k$ if $k\not=i,j$. Since $h_{(a,b)} = h_{(a+1,b-1)} + s_{(a,b)}$ whenever $a \ge b$, we have

$$h_{\mu^\star} = \bigl(\prod_{k\not=i,j} h_{\mu_k} \bigr) h_{\mu_i-1}h_{\mu_j+1}= \bigl(\prod_{k\not=i,j}h_{\mu_k}\bigr) \bigl( h_{\mu_i}h_{\mu_j} + s_{(\mu_i-1,\mu_j+1)} \bigr) $$

and so if $f = \prod_{k\not=i,j}h_{\mu_k}$ then

$$K_{\lambda\mu^\star} = \langle s_\lambda, h_{\mu^\star} \rangle = \langle s_\lambda, h_\mu \rangle + \langle s_\lambda, f s_{(\mu_i-1,\mu_j+1)} \rangle = K_{\lambda\mu} + \langle s_\lambda, f s_{(\mu_i-1,\mu_j+1)} \rangle \ge K_{\lambda\mu}.$$

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