= Solution
Use units in which the proton <electric charge> is one. The <up quark> and <down quark> both have <spin angular momentum> $1/2$ and form a flavour <isospin> doublet:
$$
\boxed{u:\ Q=\tfrac23,\ I=\tfrac12,\ I_3=\tfrac12;\qquad d:\ Q=-\tfrac13,\ I=\tfrac12,\ I_3=-\tfrac12.}
$$
Their <antiquarks> have the opposite charges and third components of <isospin>, with the same total <isospin> and spin. This is strong-interaction flavour <isospin>, not the chiral <weak isospin> assignment.
The <Clebsch-Gordan decomposition for SU2> says that spins $j_1,j_2$ combine into every $j$ from $|j_1-j_2|$ to $j_1+j_2$ in unit steps. It applies independently to spin and <isospin>. Thus
$$
\tfrac12\otimes\tfrac12=0\oplus1,\qquad \tfrac12\otimes\tfrac12\otimes\tfrac12=\tfrac32\oplus\tfrac12\oplus\tfrac12.
$$
Before exchange symmetry is imposed, three light <quarks> have total spin $S=1/2$ or $3/2$, and total <isospin> $I=1/2$ or $3/2$; the doublet occurs twice in the uncoupled tensor cube. With zero <orbital angular momentum>, $J=S$.
Physical <baryons> must also be <colour singlets>. The <three-quark colour singlet> is completely antisymmetric, so the <Pauli exclusion principle> requires symmetric spatial-spin-flavour symmetry. For a symmetric spatial S-wave state, the spin-flavour factor is symmetric. The $I=3/2$ flavour quartet is symmetric, whereas the $I=1/2$ flavour doublets have mixed permutation symmetry; the spin factors have the analogous symmetries. Symmetric flavour therefore pairs with symmetric spin, while the two mixed factors pair in their unique symmetric combination. This is the <Pauli constraint on three-quark flavour multiplets>; it leaves
$$
\boxed{(I,J)=(\tfrac12,\tfrac12)\quad\text{and}\quad(\tfrac32,\tfrac32).}
$$
Equivalently the symmetric cube of the four-dimensional two-flavour spin space has dimension $\binom63=20=2\cdot2+4\cdot4$. The unmatched spin-isospin pairings in the bare tensor product do not describe the usual symmetric-spatial ground-state <baryons>.
For three light <quarks>, charge addition gives $Q=I_3+1/2$. The observed <nucleon> doublet and <Delta baryon> quartet have the following assignments; the listed contents specify flavour counts, not the complete spin-flavour wavefunction:
$$
\begin{array}{c|c|c|c|c}
\text{state}&\text{content}&I&I_3&Q\\\hline
p&uud&1/2&1/2&1\\
n&udd&1/2&-1/2&0\\
\Delta^{++}&uuu&3/2&3/2&2\\
\Delta^+&uud&3/2&1/2&1\\
\Delta^0&udd&3/2&-1/2&0\\
\Delta^-&ddd&3/2&-3/2&-1
\end{array}
$$
The <proton> and <neutron> have $J=1/2$; all four <Delta baryons> have $J=3/2$. These states have positive <parity> in the zero-orbital ground-state model.
For a light <quark>-<antiquark> <meson>, both spin and flavour doublets combine into singlet plus triplet. In a conventional choice of phases the flavour states are
$$
\begin{array}{c|c|c}
(I,I_3)&\text{flavour state}&Q\\\hline
(1,1)&u\overline d&1\\
(1,0)&(u\overline u-d\overline d)/\sqrt2&0\\
(1,-1)&d\overline u&-1\\
(0,0)&(u\overline u+d\overline d)/\sqrt2&0
\end{array}
$$
Each row can have $J=0$ or $1$, since the two constituent spins couple to zero or one. At zero <orbital angular momentum> their <parity> is negative, following the <meson parity and charge conjugation> rule $P=(-1)^{L+1}$. The spin-zero isotriplet is the <pion> multiplet, and the spin-one isotriplet the <rho meson> multiplet. The nonstrange spin-one isosinglet is the idealized <omega meson> flavour direction. Its spin-zero counterpart is the nonstrange direction contributing to the <eta and eta prime mesons>; it is not automatically a physical mass eigenstate once other flavours and mixing are included.
Now introduce an <isospin>-zero <quark> $h$ of spin $1/2$ and charge $q_h$. For <baryons with one isospin-singlet quark>, the two light <quarks> have either the antisymmetric flavour singlet $[ud]=(ud-du)/\sqrt2$ or the symmetric triplet $uu,\{ud\},dd$. Their colour pair is antisymmetric. Hence <light-diquark spin-isospin correlation> gives $S_{ud}=0$ for $I=0$ and $S_{ud}=1$ for $I=1$. Coupling the $h$ spin produces
$$
\begin{array}{c|c|c|c}
\text{light pair}&I&J&Q\\\hline
[ud]&0&1/2&q_h+1/3\\
uu&1&1/2\text{ or }3/2&q_h+4/3\\
\{ud\}&1&1/2\text{ or }3/2&q_h+1/3\\
dd&1&1/2\text{ or }3/2&q_h-2/3
\end{array}
$$
The three triplet rows have $I_3=1,0,-1$. For a strange-like charge $q_h=-1/3$, these are the neutral spin-one-half <Lambda baryon>, the spin-one-half <Sigma baryon> triplet of charges $1,0,-1$, and its spin-three-half starred partners. For a charm-like charge $q_h=2/3$, they are the spin-one-half <Lambda c baryon> of charge $1$, and the <Sigma c baryon> triplets of charges $2,1,0$ with spins $1/2$ and $3/2$. Both possible new charge assignments have thus been covered; charge-conjugate <antibaryons> have opposite charges and unchanged $I,J$.
The <meson doublets with one isospin-singlet quark> are $h\overline u,h\overline d$, with
$$
\boxed{I=\tfrac12,\quad I_3=(-\tfrac12,\tfrac12),\quad Q=(q_h-\tfrac23,q_h+\tfrac13),\quad J=0\text{ or }1.}
$$
For $q_h=-1/3$ their charges are $-1,0$, giving $K^-,\overline K^0$-like <kaons> and their vector partners; the conjugate doublet has $u\overline h,d\overline h$ and charges $1,0$. For $q_h=2/3$ the charges are $0,1$, giving $D^0,D^+$-like <D mesons> and their vector partners, with charge-conjugate partners as well. Finally a hidden-flavour pair $h\overline h$, if included, is neutral with $I=0$ and $J=0$ or $1$. These are quantum-number classifications; charge and <isospin> alone do not determine particle masses or mixing angles.
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