For every half-integer $j=n/2,n\in\mathbb{Z}$, there is an irreducible representation of $SU(2)$
$$D^j=\exp(-i\vec\theta\cdot\vec J^{(j)})$$
in which the three generators $J_i^{(j)}$ are $(2j+1)\times(2j+1)$ square hermitian matrices. As you probably know, $D^j$ describes states with angular momentum $-j$ to $j$ in integer steps. Given two particles with angular momenta $j$ and $\ell$, we write the total angular momentum of the system as the tensor product $D^j\otimes D^\ell$. We have the standard result
$$D^j\otimes D^\ell=\bigoplus_{k=|j-\ell|}^{j+\ell}D^k$$
where on the right it is (conventionally) understood that we write the rotation matrices in order of decreasing angular momentum. Consider, for instance, a meson. This is a composite particle of two spin-half particles. In its ground state, the meson spin representation is
$$D^{1/2}\otimes D^{1/2}=D^1\oplus D^0$$
We can thus predict that there are spin-1 and spin-0 mesons, which has been experimentally verified.
Suppose then we wish to find the angular momentum of a baryon. We need $D^{1/2}\otimes D^{1/2}\otimes D^{1/2}$. Convince yourself of the following: Given three matrices $A,B,C$ we have
$$A\otimes(B\oplus C)=A\otimes B\oplus A\otimes C$$
Using this, we have
$$D^{1/2}\otimes D^{1/2}\otimes D^{1/2}=D^{1/2}\otimes(D^1\oplus D^0)=D^{1/2}\otimes D^1\oplus D^{1/2}\otimes D^0=D^{3/2}\oplus D^{1/2}\oplus D^{1/2}$$
Again, we find baryons with identical quark content in both $3/2$ and $1/2$ states. (There is a slight technicality with the second $D^{1/2}$ involving the Pauli principle.)
The obvious generalization of this is
$$D^m\otimes D^j\otimes D^\ell=D^m\otimes\left(\bigoplus_{k=|j-\ell|}^{j+\ell}D^k\right)=\bigoplus_{k=|j-\ell|}^{j+\ell}D^m\otimes D^k=\bigoplus_{k=|j-\ell|}^{j+\ell}\bigoplus_{n=|m-k|}^{m+k}D^n$$
and so on for more particles. From here you can use the Clebsch-Gordan method to construct the actual state vectors for your system.
