I would like to ask how do you count the number of possible quark combination that could possibly exist in a baryon. I know certain spin symmetry or orbital momentum symmetry have to be conserved. But I'm not quite sure how? e.g. For S=1/2 on page .5 at the bottom, why is there a minus in the second term?
The coefficients you see on page 5 (including the minus sign) are simply the usual Clebsh-Gordan coefficients. When two angular momenta $\vec{J_1}$ and $\vec{J_2}$ are added, the new space is the tensorial product $\mathcal{H}_1(j_1) \otimes \mathcal{H}_2(j_2)$ of dimension $(2j_1+1)\times(2j_2+1)$ for which a complete basis can be denoted by the following eigenvectors: \begin{equation*} |j_1, j_2, m_1, m_2> \equiv |j_1,m_1> \otimes |j_2,m_2> \end{equation*} These states are eigenvectors of $\hat{J}_1^2, \hat{J}_{1_z}, \hat{J}_2^2, \hat{J}_{2_z}$ One can show that there exists another complete basis of this Hilbert space which corresponds to the total angular momentum $\vec{J}=\vec{J_1}+\vec{J_2}$. This basis is generated by the eigenvectors of $J^2_1, J^2_2, J^2$, and $J_z$: \begin{equation*} |j_1,j_2,j,m> ~\mbox{with} ~ j \in j_1+j_2,~j_1+j_2-1,~\cdots,~ |j_1-j_2|~\mbox{and}~m = m_1+m_2 \in [-j,j] \end{equation*} In other words, the resulting space is decomposed as: \begin{equation*} \mathcal{H}_1(j_1) \bigotimes \mathcal{H}_2(j_2) = \bigoplus_{j=|j_1-j_2|}^{j_1+j_2} \mathcal{H}(j) \end{equation*} The relationship between the basis $|j_1,j_2j,m>$ and $|j_1, j_2, m_1, m_2>$ is given by the Clebsch-Gordan coefficients, chosen as real numbers and denoted below as $<j_1, j_2, m_1, m_2|j_1,j_2,j,m>$: \begin{equation*} |j_1,j_2,j,m> = \sum_{m_1=-j_1}^{j_1} ~\sum_{m_2=-j_2}^{j_2} <j_1, j_2, m_1, m_2|j_1,j_2,j,m>|j_1, j_2, m_1, m_2> \end{equation*} where only coefficients for which $m = m_1+m_2$ are non-zero. The most useful coefficients for particle physics are given in a concise way in this figure: http://pdg.lbl.gov/2014/reviews/rpp2014-rev-clebsch-gordan-coefs.pdf
