= Solution
Compared with the complete three-flavour <meson> <tensor product>, the additional flavour states are exactly
$$
\boxed{\underbrace{(u\bar c,\ d\bar c,\ s\bar c)}_{\mathbf3},\qquad
\underbrace{(c\bar u,\ c\bar d,\ c\bar s)}_{\overline{\mathbf3}},\qquad
\underbrace{c\bar c}_{\mathbf1}.}
$$
Thus there are seven new flavour states. For the <pseudoscalar mesons>, the six open-charm states are $\bar D^0,D^-,D_s^-$ and $D^0,D^+,D_s^+$, respectively, using the <D meson> and <Ds meson> conventions. A pure $c\bar c$ <pseudoscalar meson> is the idealized flavour state associated with the <eta c meson>. The <quark model> flavour calculation alone does not specify spin or spatial excitations, so the same flavour labels also organize other <meson> species.
If comparing the traceless <SU(4)> $\mathbf{15}$ with the traceless <SU(3)> <meson octet>, the extra seven states are instead the same two triplets and the normalized <SU(3)> singlet
$$
\boxed{\eta_{15}=\frac{u\bar u+d\bar d+s\bar s-3c\bar c}{\sqrt{12}}.}
$$
Its coefficient <matrix> is traceless and has squared norm one. In particular, $c\bar c$ by itself is not a vector in the traceless $\mathbf{15}$; its coefficient <matrix> has <trace> one.
To relate these two descriptions, let $S_4=(u\bar u+d\bar d+s\bar s+c\bar c)/2$ be the normalized <SU(4)> singlet and $S_3=(u\bar u+d\bar d+s\bar s)/\sqrt3$ the old <SU(3)> singlet. Then
$$
\boxed{S_3=\frac{\sqrt3}{2}S_4+\frac12\eta_{15},\qquad
c\bar c=\frac12S_4-\frac{\sqrt3}{2}\eta_{15}.}
$$
This orthonormal change of <basis> shows explicitly why the complete <tensor product> contains two light-flavour singlets but adds only one state to the old light-flavour singlet sector.
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