QFT - Allowed transitions for a given vertex

I’m given the following Lagrangian :

$$ L = L_{f1 }(psi_1, barpsi_1) + L_{f2}(psi_2, barpsi_2) + lambda (barpsi_1 gamma^{mu} psi_2) (barpsi_2 gamma_{mu} psi_1) $$

where $L_{fi}$ are two different free fermionic Lagrangian.
What I understand, just by looking at the form of the interaction term, is that we have just one type of vertex, that involves 4 different fields: $psi_1, psi_2, barpsi_1, barpsi_2$. Limiting our discussion to $2 to 2$ processes, my guess is that the only interactions, with a non null first order contribution, would have the form $ (psi_1 psi_2 to barpsi_1 barpsi_2)$, $(barpsi_1 barpsi_2 to psi_1 psi_2)$ or $ (psi_i barpsi_j to psi_j barpsi_i)$; in other terms I’m considering all the possible permutations of the 4 listed fields.

But then, looking at my professor’s notes, he computes the total amplitude of scattering processes — like $(psi_1 psi_2 to psi_1 psi_2)$ or $(barpsi_i psi_j to barpsi_i psi_j)$ — simply using the single vertex rule (i.e. like a first order process). Why is so? Is that just a coincidence (maybe an error) or a consequence of the simmetry of the given Lagrangian? In the case he meant to consider a first order computation, why can we use the same vertex rule, arbitrarily replacing the fields with their complex/dagged/barred version?

I have an extra question: if I’m considering a $3 to 1$ (or, equivalently a $1 to 3$) process, can I just use the same vertex rule? I’m propending to an affermative answer, since the vertex doesn’t distinguish (in this case) between incoming and outgoing particles.