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A skew Young diagram $\lambda /\mu$ is the difference of two nested Young diagrams. A semistandard Young tableau is a filling of precisely those cells by positive integers, weakly increasing along each occupied row and strictly increasing down each occupied column. It can be viewed as its filled diagram, or equivalently as a chainwhere ${\nu }^{(r)}/{\nu }^{(r-1)}$ is a horizontal strip consisting of the cells labeled $r$. Indeed, the weak row and strict column inequalities ensure that the cells labeled at most $r$, together with $\mu$, form a partition diagram, and equal labels cannot share a column. Conversely such a chain gives the tableau.

For a tableau $T$, let $c_r(T)$ count its entries $r$, put $x^T={\prod }_r{x}_r^{c_r(T)}$, and defineThere are finitely many tableaux for any fixed number of allowed letters, and the sum has total degree $|\lambda |-|\mu |$.

To prove symmetry directly, use a Bender-Knuth involution for each adjacent pair of letters $i,i+1$. Pair any $i$ and $i+1$ that occur in the same column and leave them fixed; strictness makes them consecutive in that column. In a row the remaining unpaired entries have the form $i^a(i+1{)}^b$, between any fixed paired letters. Replace that block by $i^b(i+1{)}^a$. Rows stay weakly ordered. A free $i$ can be changed to $i+1$ without equaling the entry below, because such a below entry would have been paired with it; likewise changing a free $i+1$ to $i$ cannot equal the entry above. Other neighboring letters are outside $\{i,i+1\}$, so strict columns are preserved. The paired cells remain the same, and repeating the operation restores each free block. The paired letters contribute equally to the two contents, while every free block exchanges its contributions. Thus the involution is a weight-preserving bijection after exchanging $x_i,x_{i+1}$. Adjacent exchanges generate all finite variable permutations, provingThe skew Jacobi–Trudi identity gives a second proof, since its determinant entries are already symmetric.

Define the Kostka number $K_{\lambda \mu }$ to count straight-shape Semistandard Young tableaux with shape $\lambda$ and content $\mu$, meaning ${\mu }_i$ copies of label $i$. Suppose such a tableau exists. Strict columns imply that an entry in row $r$ is at least $r$. Therefore all entries from $1,\dots ,r$ lie in the first $r$ rows, givingHenceFor content $\lambda$, the ${\lambda }_1$ copies of $1$ have nowhere to go except row one and fill it. The ${\lambda }_2$ copies of $2$ then fill row two, and induction fills row $r$ with $r$. This filling exists and is unique, so $\boxed{K_{\lambda \lambda }=1}$.

Symmetry identifies equal coefficients for each rearrangement of a content, henceThe monomial symmetric functions form a basis. In any total order extending dominance, this change-of-basis matrix is triangular with diagonal one. It is invertible in each degree, so

Sum the weight monomials of all Semistandard Young tableaux on the skew Young diagram $\lambda /\mu$, with weak rows and strict columns. A Bender-Knuth involution interchanges the counts of adjacent letters without changing the shape, so the sum is a symmetric function. Taking $\mu$ empty gives the Schur function.