I am referring to this, http://home.web.cern.ch/about/updates/2014/04/lhcb-confirms-existence-exotic-hadron
So how does this work if we stick to keeping quarks in the 3 dimensional fundamental representation of $SU(3)$?
This bound-state seems to have 2 anti-quarks and 2 quarks. So with just 3 colours how do we make the whole thing anti-symmetric with respect to the colour quantum number?
Is there anything called "anti-colour" quantum number that an anti-quark can posses so that there are a total of $(3\times 2)^2$ colour options to choose from for the 2 quarks and 2 anti-quarks? I have never heard of such a thing!
The point is that unlike the $U(1)$ charge, the non-Abelian charge doesn't occur in the Lagrangian for the quarks. The Lagrangian only sees the different flavours, the gauge groups and the gauge coupling constant.