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 $\lambda \unrhd \mu$.



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


$$K\_{\lambda\mu^\star} = \langle s\_\lambda, h\_{\mu^\star} \rangle = \langle s\_\lambda, h\_\mu \rangle + \langle s\_\lambda, f s\_{(\mu\_i-1,\mu\_j+1)} \rangle = K\_{\lambda\mu} + \langle s\_\lambda, f s\_{(\mu\_i-1,\mu\_j+1)} \rangle \ge K\_{\lambda\mu}.$$



Is there a short proof that the Kostka number $K_{\lambda \mu}$ is non-zero whenever $\lambda$ dominates $\mu$?
I think the following is a simple combinatorial argument which constructs the most dominant semistandard $\lambda$-tableau of content $\mu$ whenever $\lambda\trianglerighteq\mu$. (n.b. I haven't followed the reference given in Richard Stanley's comment, so I don't know whether I'm duplicating what's done there.)


In a nutshell, the idea is "put the largest numbers as low as possible". So let $l$ be the length of $\mu$, and start the tableau by putting the $\mu\_l$ $l$s in the bottom boxes of columns as far to the left as possible subject to the condition that if any box has an $l$ and there is a box directly to the right, then that box must also have an $l$ in it. (Another way of saying this is: put $l$s at the bottom of the first $\mu\_l$ columns, and then slide these $l$s to the ends of their rows.) If we let $j$ be maximal such that $\lambda\_j\geq\mu\_l$, this means that we are putting $\lambda\_x-\lambda\_{x+1}$ $l$s at the end of row $x$ for each $x>j$, and $\mu\_l-\lambda\_{j+1}$ $l$s at the end of row $j$.


To fill in the rest of the tableau, we work recursively. Let $\hat\lambda$ denote the partition whose Young diagram comprises the boxes that are still empty, and let $\hat\mu$ be the partition $(\mu\_1,\dots,\mu\_{l-1})$. Then as long as $\hat\lambda\trianglerighteq\hat\mu$, we can fill in the rest of the tableau with a semistandard $\hat\lambda$-tableau of content $\hat\mu$, which gives us what we need.


So we need to show that $\hat\lambda\_1+\dots+\hat\lambda\_x\geq\hat\mu\_1+\dots+\hat\mu\_x$ for every $x$. For $x<j$ or $x\geq l$ this is immediate from the fact that $\lambda\trianglerighteq\mu$, so take $j\leq x<l$. Observe that $$\lambda\_1+\dots+\lambda\_x=n-(\lambda\_{x+1}+\dots+\lambda\_l)\geqslant n-(l-x)\lambda\_{x+1}$$while $$\mu\_1+\dots+\mu\_x=n-(\mu\_{x+1}+\dots+\mu\_l)\leqslant n-(l-x)\mu\_l.$$ So $$(\hat\lambda\_1+\dots+\hat\lambda\_x)-(\hat\mu\_1+\dots+\hat\mu\_x)=(\lambda\_1+\dots+\lambda\_{x+1}-\mu\_l)-(\mu\_1+\dots+\mu\_x)\geqslant(l-x-1)(\mu\_l-\lambda\_{x+1})\geqslant0$$ as required.


