The total wave function for a particle is made by a combination of a spin part (say $\chi$) and of a position part (lets say $\psi$), so that we have :
$$
\Psi_t(x_i,s_i)=\chi (s_i) \, \psi (x_i)
$$
When you consider bosons, you want to ensure that the total wave function is symmetric. As such, you would have either two cases: a symmetric spin wave function and a symmetric spatial wave function or you would have that the spin and spatial wave functions are both anti-symmetric.
So, for bosons (+ sign for symmetric and - for anti-symmetric):
$$
\Psi_t = \chi_+ \ \psi_+
$$
Or
$$
\Psi_t = \chi_- \ \psi_-
$$
Your final wave function is then a combination or those two cases.
For fermions, same idea but remember that fermions have anti-symmetric total wave function. So that for fermions, both accepted wave functions are:
$$
\Psi_t = \chi_- \ \psi_+
$$
Or
$$
\Psi_t = \chi_+ \ \psi_-
$$
Now say you consider a spin $\frac{1}{2}$ particle, there are four spin states, which we separate into two categories, the singlet state and the triplet states.
For the singlet state (anti-symmetric state), you have:
$$
|\chi\rangle = \frac{1}{\sqrt{2}}[|\uparrow\downarrow\rangle - |\downarrow\uparrow\rangle]
$$
As for the triplet states (all symmetric states), they are :
$$
|\chi\rangle = \frac{1}{\sqrt{2}}[|\uparrow\downarrow\rangle + |\downarrow\uparrow\rangle] \\ |\chi\rangle= |\uparrow\uparrow\rangle \\|\chi\rangle= |\downarrow\downarrow\rangle
$$
You can then see that it is possible to get extra degeneracy via the spin parts (you have more than one state that gives a symmetric spin part).
For extra reference, you can check out Sakurai's or Zettili's textbook.
