 Solution

A meson has quark model flavour content $q_i{\bar{q}}^j$, so four quark flavours give sixteen flavour states inThis counts a flavour set for a specified spin and orbital assignment; it does not count every excited meson. Three quark flavours instead give the familiar nine states $3\otimes \bar{3}=8\oplus 1$.

Taking the fourth quark to be the charm quark, the first nine combinations $q_{\alpha }{\bar{q}}^{\beta }$, with $\alpha ,\beta \in \{u,d,s\}$, are the light sector. The extra seven are $q_{\alpha }\bar{c}$ in $3$, $c{\bar{q}}^{\alpha }$ in $\bar{3}$, and $c\bar{c}$ in $1$ under light-flavour $SU(3)$. For the lowest pseudoscalar mesons, the open-charm states are the three D meson/Ds meson states and their antiparticles, and the hidden-charm state is the eta c meson. The two neutral trivial representation states can mix when the flavour symmetry is broken.

The intended limitation is the accuracy of flavour symmetry: the charm quark is much heavier than the three light quarks, so a quark mass term is far from $SU(4)$ invariant. Exact flavour $SU(4)$ would relate light and charmed states with the same spin, contrary to their large observed mass splittings. However, saying that a four-flavour quark model is never considered would be too strong: flavour $SU(4)$ is used as classification bookkeeping, with much stronger symmetry breaking than light-flavour $SU(3)$.^[ref] This is a flavour special unitary group, distinct from the three-colour gauge group of Quantum chromodynamics.