The Fourier expansion of the fermion field operator is such that
$$ \hat\psi=\int\!d^3p\,\left[ f_b(p)\hat b(p) +f_d(p)\hat d^\dagger\!(p) \right] ~~, $$
for some sufficiently complicated $f_b$ and $f_d$. The operators $\hat b^\dagger$ and $\hat b$ create and destroy electrons respectively, and $\hat d^\dagger,\,\hat d$ work the same for positrons. I haven't seen it stated explicitly, but I assume both $b$ operators commute with each $d$ operator. What is an efficient way to show, for instance,
$$ \hat b^\dagger\hat d^\dagger\big|0\big\rangle= \hat d^\dagger\hat b^\dagger\big|0\big\rangle~~? $$
Any demonstration that $b$ and $d$ commute would suffice if another is more direct, $[\hat b,\hat d]=0$ for instance.