= Solution
The diagram refers to <ground state>, $L=0$, <color singlet> <baryons>. Their spatial <wavefunction> is symmetric and their color <wavefunction> is antisymmetric. The <Pauli exclusion principle> therefore requires a symmetric combined spin-and-flavor <wavefunction>. One cannot independently attach every three-spin coupling to each flavor assignment.
For three spin-$1/2$ particles, there are formally two ways to couple to total <spin> $1/2$. In the <ground state> <quark model> these are tied to the corresponding mixed-symmetry flavor functions. Their required symmetric combination yields one physical spin-$1/2$ flavor octet, not two independent species at each peripheral weight. In <spin-flavor symmetry> notation,
$$
\mathbf{56}_{SU(6)}=(\mathbf{10},\mathbf4)\oplus(\mathbf8,\mathbf2)
$$
under flavor $SU(3)$ and spin $SU(2)$. Thus there is one peripheral species in the <baryon octet>, each still possessing its two ordinary spin projections. The two abstract flavor-octet copies in $\mathbf3\otimes\mathbf3\otimes\mathbf3$ do not double the physical <ground state> octet.
At the dotted positions, the flavor content would be $uuu$, $ddd$ or $sss$. These flavor <wavefunctions> are completely symmetric. The spin <wavefunction> must then also be completely symmetric, which means total <spin> $3/2$, not $1/2$. The corresponding <baryon decuplet> states are $\Delta^{++}$, $\Delta^-$ and $\Omega^-$. This exclusion is for the stated <ground state> spin-$1/2$ multiplet; it is not an exclusion of all excited <baryons> at those weights.
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