Normal Spinel and Inverse Spinel
Let's take a (2-3) spinel ($\ce{A^{II}B^{III}_{2}O4}$) as an example. If it is normal or inverse can be predicted from the total crystal field stabilization energies. For a normal spinel, the total CFSE wold be: $$\Delta_{T}(A)+2\Delta_{O}(B)=\Delta_{T}(A)+\Delta_{O}(B)+\Delta_{O}(B)$$ and for an inverse spinel $$\Delta_{O}(A)+\Delta_{T}(B)+\Delta_{O}(B)$$ with $\Delta_{T}$ being the tetrahedral and $\Delta_{O}$ being the octahedral splitting energy of $\ce{A^2+}$ and $\ce{B^3+}$, respectively. 

If $\Delta_{T}(A)+\Delta_{O}(B)&gt;\Delta_{O}(A)+\Delta_{T}(B)$, the normal spinel structure should be favored, as the resulting CFSE is higher. If it is the other way around, an inverse spinel is to be expected.

To sum it up, if the CFSE gained by placing $\ce{A^2+}$ in the octahedral holes outweighs a potential destabilization caused by putting a part of $\ce{B^3+}$ in the tetrahedral holes, an inverse spinel can be formed. However, a normal spinel is favorable when $\ce{B^3+}$ clearly "prefers" the octahedral and/or $\ce{A^2+}$ the tetrahedral holes.