= Solution
A <meson> has <quark model> flavour content $q_i\bar q^{\,j}$, so four <quark flavours> give sixteen flavour states in
$$
\boxed{\mathbf4\otimes\bar{\mathbf4}=\mathbf{15}\oplus\mathbf1.}
$$
This 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 $\mathbf3\otimes\bar{\mathbf3}=\mathbf8\oplus\mathbf1$.
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 $\mathbf3$, $c\bar q^{\,\alpha}$ in $\bar{\mathbf3}$, and $c\bar c$ in $\mathbf1$ 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 $SU(3)$.<Quark Model in the 2024 Review of Particle Physics>[Section 15.5.2, printed p. 19: the large charm mass strongly breaks flavour SU(4), which remains useful for bookkeeping]{ref} This is a flavour <special unitary group>, distinct from the three-colour <gauge group> of <Quantum chromodynamics>.
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