A disconnected removable rim set is not a border strip

The cells of $[\lambda ]$ are $(1,1),(1,2),(2,1),(2,2),(3,1),(4,1)$, those of $[\nu ]$ are $(1,1),(2,1),(3,1)$, and $\nu \subseteq \lambda$; hence $\lambda \mathrm{/}\nu =\{(1,2),(2,2),(4,1)\}$ with $r=6-3=3$ nodes, and removing these nodes from $[\lambda ]$ leaves $[\nu ]$, a Young diagram. The set is the union of the two skew hooks $\{(1,2),(2,2)\}$ (a vertical domino in column $2$) and $\{(4,1)\}$ (a single node), which occupy the $c=2$ columns $1$ and $2$ and are pairwise disjoint, so $m=2$. The two components are not edge-adjacent: the only side-neighbour of $(2,2)$ in column $2$ below it is $(3,2)\mathrm{\notin }[\lambda ]$, and $(4,1)$ has no side-neighbour in $\lambda \mathrm{/}\nu$. Hence $\lambda \mathrm{/}\nu$ is disconnected and, by [F1], is not a border strip of $\lambda$, and in particular is not a removable border strip of $\lambda$.

The Littlewood–Richardson coefficients $c_{(1,1,1),\mu }^{(2,2,1,1)}$ are $1,1,0$ for $\mu =(1,1,1),(2,1),(3)$: the reading order of the cells is $(1,2),(2,2),(4,1)$; for content $(1,1,1)$ a lattice word with three distinct labels must read $1,2,3$, forcing the entries $1,2,3$ in that order down the cells $(1,2),(2,2),(4,1)$, which respects the strict increase of column $2$ and is the unique such tableau; for content $(2,1)$ the lattice words are $112$ and $121$, and only $121$ has the first two entries strictly increasing, giving the unique tableau $(1,2;1)$; for content $(3)$ the word $111$ puts equal labels $1,1$ in the two cells of column $2$, violating strict increase, so there is none. In the binomial form of [F3] these are $\left(\genfrac{}{}{0px}{}{1}{2-x}\right)$, namely $\left(\genfrac{}{}{0px}{}{1}{1}\right)=1$, $\left(\genfrac{}{}{0px}{}{1}{0}\right)=1$, $\left(\genfrac{}{}{0px}{}{1}{-1}\right)=0$ for $x=1,2,3$.

Therefore ${\sum }_{x=1}^3(-1{)}^{3-x}{c}_{(1,1,1),(x,1^{3-x})}^{(2,2,1,1)}=(+1)\cdot 1+(-1)\cdot 1+(+1)\cdot 0=0$, in accordance with the $m\ge 2$ case of [F3], even though the coefficients at $x=1$ and $x=2$ are both nonzero; the disconnected set contributes the nonzero terms $1$ and $1$, which cancel in the alternating sum, so it cannot be assigned a single sign $(-1{)}^{\mathrm{ht}}$ as a recursion summand.

Consistently, the hook lengths of $\lambda$ are $5,4,2,2,1,1$ [F4], none equal to $3$, so ${\chi }^{(2,2,1,1)}(\sigma )=0$ for every $\sigma \in S_6$ whose cycle type contains a part $3$; there is no removable border strip of size $3$, although step 1.1 exhibits a disconnected removable set of that size. To test clause 2 numerically, extend the proposed height to mean occupied rows minus one and evaluate at $(3,1,1,1)$. The only partitions of $3$ contained in $\lambda$ are $(2,1)$ and $(1,1,1)$, and both differences are disconnected, $2\times 2$-free sets occupying three rows. Their proposed signs are both $+1$, so the erroneous sum is ${\chi }^{(2,1)}(1^3)+{\chi }^{(1,1,1)}(1^3)=2+1=3$, using the hook formula (The hook length formula) with hook products $3$ and $6$, respectively. This differs from the correct value $0$.

Step 1.1 exhibits a skew diagram $\lambda \mathrm{/}\nu$ that is a removable union of $m=2$ disjoint skew hooks with $3$ nodes but is not edge-connected, refuting clause 1; steps 1.2 and 2.1 show that the individual Littlewood–Richardson coefficients of this disconnected union are nonzero ($1$ and $1$) while their alternating sum vanishes, and step 2.2 computes the wrong value $3$ instead of $0$ for the proposed signed recursion, refuting clause 2; only the connected border strips contribute $(-1{)}^{r-c}$ to the alternating sum, exactly as in [F3].