Counting states for $\ell=3$ Plus spin ${1\over 2}$

For $\ell=3$ there are $2\ell+1=7$ different eigenstates of $L_z$. There are two different eigenstates of $S_z$ for spin ${1\over 2}$. We can have any combination of these states, implying $2\times 7=14$ possible product states like $Y_{31}\chi_+$.

We will argue based on adding vectors... that there will be two total angular momentum states that can be made up from the 14 product states, $j=\ell\pm{1\over 2}$, in this case $j={5\over 2}$ and $j={7\over 2}$. Each of these has $2j+1$ states, that is 6 and 8 states respectively. Since $6+8=14$ this gives us the right number of states.