In a compound oxide ions have CCP arrangement. Cations  $A$ are present in one-eighth of the tetrahedral holes and cations $B$occupy half of the octahedral holes. The simplest formula of the compound is

In cubic close packing, also known as CCP, the number of atoms per unit cell is 4.

Therefore the number of oxide present per unit cell $=4$

So, the number of tetrahedral voids $=2n =2\times 4=8$

Similarly, the number of octahedral voids $=\frac{1}{8}\times 8=1$

Finally, the number of cations $B$ per unit cell $=\frac{1}{2}\times 4=2$

Hence, the formula for the compound will be in ratio  $A:B:O^{2-}=1:2:4$  that is $A{B}_2{O}_4$.