A fermion is a particle that follows Fermi–Dirac statistics and generally has half odd integer spin $1/2, 3/2$, etc. On the other hand, a boson is a particle that follows Bose-Einstein statistics and generally has integer spin $0,1,2$, etc.
In general, the wave function is equal to the product of the partial part and the spin part
$$\Psi(\boldsymbol{r},\boldsymbol{S})=\psi(\boldsymbol{r})\chi(\boldsymbol{S})$$
The consequence of the fermions is that the wave function of the particle must be anti-symmetric while for bosons the wave function of the particle must be symmetric.
Thus for fermions :
$$ \Psi(\boldsymbol{r},\boldsymbol{S}) = \begin{cases}
\psi_a(\boldsymbol{r})\cdot \chi_s(\boldsymbol{S}) & \\
\psi_s(\boldsymbol{r})\cdot \chi_a(\boldsymbol{S}) &
\end{cases}
$$
for bosons :
$$ \Psi(\boldsymbol{r},\boldsymbol{S}) = \begin{cases}
\psi_a(\boldsymbol{r})\cdot \chi_a(\boldsymbol{S}) & \\
\psi_s(\boldsymbol{r})\cdot \chi_s(\boldsymbol{S}) &
\end{cases}
$$
Edit: I'm adding a simple example for two non-interacting particles.
The symmetric and antisymmetric parts are used as follows:
$$\psi_a=\psi_{n_1}(\mathbf{r}_1)\psi_{n_2}(\mathbf{r}_2)-\psi_{n_1}(\mathbf{r}_2)\psi_{n_2}(\mathbf{r}_1)$$
$$\psi_s=\psi_{n_1}(\mathbf{r}_1)\psi_{n_2}(\mathbf{r}_2)+\psi_{n_1}(\mathbf{r}_2)\psi_{n_2}(\mathbf{r}_1)$$
First If the particles are bosons then the
$$\Psi= \begin{cases}
\psi_a(\boldsymbol{r})\cdot \chi_a(\boldsymbol{S}) & \\
\psi_s(\boldsymbol{r})\cdot \chi_s(\boldsymbol{S}) &
\end{cases}$$
for the fermions
$$\Psi= \begin{cases}
\psi_a(\boldsymbol{r})\cdot \chi_s(\boldsymbol{S}) & \\
\psi_s(\boldsymbol{r})\cdot \chi_a(\boldsymbol{S}) &
\end{cases}$$
If the fermion is happen to be spin-$1/2$
particle like electron you can use $\chi_a$ to be triplet and $\chi_s$ to be singlet.
Reference
- Identical Particles : Quantum Mechanics Concepts and Applications :-Nouredine Zettili