Permitted bound states in extended QCD color group $SU(4)$ I'm very much curious about the possible quark combinations to form bound states in $SU(4)$ QCD.
We already know Wilson has proven that only colorless combinations can arise in $SU(3)$ of these formats: $q\bar{q}$, $qqq$, tetra quarks and penta quarks.
I guess in $SU(4)$ one can only find $q\bar{q}$ and $qqqq$ and combinations of these two.
Am I right? 
 A: In fact yes, a simple way to get all possible states is by the following argument: quarks belong to the fundamental representation of the color group, and to them one can assign the upper index $q^{i}$, the antiquarks belong to the antifundamental representation, and to them we assign the lower index $\bar{q}_i$. Also there is a Levi-Civita symbol $\varepsilon^{i_1 \ldots i_N}$  ($N$ being the number of colors). The asymptotic states in QCD are color singlets $-$ so to get all possible states, you should contract all indices with each other in some way. 
As for quark-antiquark pair you form a singlet by:
$$
\bar{q}_i q^{i} \,,
$$Which means taking the trace in color space. And baryons (antibaryons) correspond to:
$$
\varepsilon_{i_1 \ldots i_N} q^{i_1} \ldots q^{i_N} \,,\qquad
\varepsilon^{i_1 \ldots i_N} \bar{q}_{i_1} \ldots \bar{q}_{i_N}
\,.$$ 
A: The argument is still the same: A hadron has to be a color singlet ${\bf 1}$ under the $SU(N)_C$ color gauge group, due to color confinement.


*

*A single quark $q$ transforms in the fundamental representation ${\bf N}$ of $SU(N)_C$, and is hence not allowed since $N>1$. 

*In a meson, the quark-antiquark-pair $q\bar{q}$ belongs to ${\bf N}\otimes\bar{\bf N}\cong{\bf 1}\oplus({\bf N^2\!-\!1})_{\rm Adj}$, which contains a singlet ${\bf 1}$, and is hence allowed. 

*In a baryon, the $N$ quarks $\overbrace{qq\ldots q}^{N\text{ times}}$ form a totally antisymmetric representation $\wedge^N {\bf N}\cong {\bf 1}$ of $SU(N)_C$, which is isomorphic to a singlet ${\bf 1}$, and is hence allowed. 
As one can see the number of quarks minus the number of anti-quarks should be divisible by $N$. For $N=3$, see also my related Phys.SE answer here.
