Why $\pi^+$ meson possible combinations of quarks is only three? I am a starter at nuclear and particles physics. I am reading particles and nuclei an introduction to the physical concepts. There was a paragraph about Colour-neutral particles and how only colourless particles can be found as free particles. Later the book adresses to the $\pi^+$ meson and states that it has only three possible combinations:
$$
\left\vert{\pi^+}\right\rangle = \left\{\begin{matrix}
\left\vert u_r\bar d_{\bar r}\right\rangle\\
\left\vert u_g\bar d_{\bar g}\right\rangle\\
\left\vert u_b\bar d_{\bar b}\right\rangle
\end{matrix}\right.
$$
My question is why they didn't count $\left\vert u_{\bar r}\bar d_r\right\rangle$ or others like that as possible. Does antiquarks can only gain anti-colours? and if so why?
 A: In quantum mechanics the operation of swapping the charge is called charge conjugation and this operation swaps a particle for its antiparticle. So for example if we take an electron and swap the charge from $-e$ to $+e$ the resulting particle is a positron. That means we cannot have an electron with a positive charge because that particle is a positron, and likewise we cannot have a positron with a negative charge because that particle is an electron.
And this applies to quarks as well. So to take your example, if we start with $u_r$ and swap the charge to $\bar r$ the result is an anti-up quark. We cannot have a "non-anti" quark with an anti-charge because that particle is by definition an anti-quark.
I guess what's confused you is the notation of putting bars on both the quark and the charge i.e. writing $\bar u_{\bar r}$ seems to suggest we could have other permutations like $u_{\bar r}$ or $\bar u_r$. But don't be mislead by this. Swapping the charge also swaps the particle to its antiparticle.
