= Solution
When the three light-quark masses are equal, the mass term is proportional to the identity in flavour space, so rotating $q=(u,d,s)$ by <SU(3)> leaves it unchanged. The gluon coupling is also flavour blind. Nearly equal masses therefore give an approximate <flavour symmetry>, distinct from the exact colour gauge symmetry. The <quark> transforms as $\mathbf3$ and the <antiquark> as $\overline{\mathbf3}$.
For <mesons>, a quark-antiquark tensor is a $3\times3$ matrix. Its trace and traceless parts are invariant subspaces, giving
$$
\boxed{\mathbf3\otimes\overline{\mathbf3}=\mathbf1\oplus\mathbf8.}
$$
The trace is the <eta singlet state>; the traceless part is the <meson octet>. For <baryons>, first separate a pair of <quarks> into symmetric and antisymmetric tensors:
$$
\mathbf3\otimes\mathbf3=\mathbf6\oplus\overline{\mathbf3}.
$$
The antisymmetric pair transforms as $\overline{\mathbf3}$ by contraction with the invariant epsilon tensor. In $\mathbf6\otimes\mathbf3$, complete symmetrization gives $\operatorname{Sym}^3\mathbf3$ of dimension $\binom53=10$; the remaining mixed-symmetry subspace has dimension eight. In $\overline{\mathbf3}\otimes\mathbf3$, trace and traceless parts give a singlet and another octet. Thus the <tensor cube of the defining SU(3) representation> is
$$
\boxed{\mathbf3^{\otimes3}=\mathbf{10}\oplus\mathbf8\oplus\mathbf8\oplus\mathbf1.}
$$
The two octets carry the two mixed permutation-symmetry components. Their <weight diagrams> are identical; the extra copy is a multiplicity, not a new arrangement of weights.
The printed charge relation fixes the <half-hypercharge convention>, rather than the usual <Flavor hypercharge> normalization:
$$
Q=I_3+Y,\qquad Y=\frac{B+S}{2}.
$$
The <quark> weights $(I_3,Y)$ are $u:(1/2,1/6)$, $d:(-1/2,1/6)$, $s:(0,-1/3)$; adding three weights gives the <baryon weight diagrams in the half-hypercharge convention>. The <baryon octet> has $(p,n)$ at $Y=1/2$, the isotriplet $(\Sigma^+,\Sigma^0,\Sigma^-)$ and the isosinglet $\Lambda^0$ at $Y=0$, and $(\Xi^0,\Xi^-)$ at $Y=-1/2$. The <baryon decuplet> has rows $\Delta$ at $Y=1/2$, $\Sigma^*$ at zero, $\Xi^*$ at $-1/2$, and $\Omega^-$ at $-1$. The three-quark flavour singlet is the antisymmetric $uds$ state at $(0,0)$.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2005/iii/paper-50-baryon-weights.png]
{title=Flavour-singlet, octet and decuplet baryon weights with Y equal to half the usual hypercharge}
The fully symmetric flavour representation is precisely the <baryon decuplet>, $\mathbf{10}=\operatorname{Sym}^3\mathbf3$. For a spatial <ground state>, the orbital wavefunction is symmetric. Spin $3/2$ is the symmetric part of three spin-$1/2$ <quarks>, so a spin-$3/2$ decuplet would have a completely symmetric space-spin-flavour wavefunction. In particular the $uuu$ state with all spins aligned has no antisymmetric factor. This conflicts with the <Pauli exclusion principle> for the three <fermions>.
An additional colour degree of freedom resolves the conflict. With three colour labels the normalized colour state is
$$
|\mathrm{colour}\rangle=\frac1{\sqrt6}\epsilon_{abc}|a\rangle|b\rangle|c\rangle.
$$
It is completely antisymmetric and an <SU(3)> colour singlet. Multiplying it by the symmetric remaining wavefunction gives the required fermionic antisymmetry. Three orthogonal labels are the minimum needed for a nonzero third <exterior power>: $\bigwedge^3\mathbb C^m=0$ for $m<3$. Thus the decuplet suggests at least three colour states, and three supply the minimal colour-singlet construction used in the <quark model>. This argument motivates three colours rather than deriving their exact number without further physical input.
The <Pauli constraint on three-quark flavour multiplets> also explains the absence of a flavour-singlet <baryon> in the lowest spatial multiplet. After extracting the antisymmetric colour factor, spin-flavour must be symmetric. Singlet flavour is completely antisymmetric, so its spin factor would also have to be completely antisymmetric. But
$$
\bigwedge^3\mathbb C^2=0,
$$
so three spin-$1/2$ <quarks> have no such spin state. By contrast, mixed flavour and mixed spin can combine symmetrically to produce the ground-state <baryon octet> with spin $1/2$. The two abstract flavour-octet copies do not imply two independent ground-state octets after the full <Pauli constraint on three-quark flavour multiplets> is applied. Excited orbital wavefunctions can permit <three-quark flavour singlet>; the exclusion is for the spatially symmetric low-energy ground multiplet, not for every <baryon> at every energy.
A flavour-singlet <meson> is allowed: a <quark> and an <antiquark> are not three identical <fermions>, and the invariant flavour combination $(u\bar u+d\bar d+s\bar s)/\sqrt3$ can combine with an appropriate spin and colour singlet. The missing three-spin antisymmetric state therefore causes no <meson> exclusion.
Finally, $\pi^+p$ has <baryon number> one, <strangeness> zero and $I_3=3/2$, so it is a pure $I=3/2$ channel. These quantum numbers match the $uuu$ <Delta baryon> $\Delta^{++}$, the low-energy resonance with spin $3/2$. For $K^+p$, the <kaon> $u\bar s$ contributes <strangeness> plus one, giving total $B=1$, $S=+1$. No ordinary $qqq$ <baryon> can have positive <strangeness>, since every $s$ <quark> contributes minus one. Hence that channel has no matching state in the ordinary low-energy three-quark multiplets. A resonance would require exotic <quark> content, for example $uuud\bar s$, and is not mandated by this spectrum. This is <strangeness selection in pion-proton and kaon-proton resonances>.
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