# Formula of the compound based on occupancy of lattice voids

Atom of element $\ce{B}$ forms HCP lattice and those of the element $\ce{A}$ occupy 2/3rd of tetrahedral voids. What is the formula of the compound formed by the elements $\ce{A}$ and $\ce{B}$?

My doubt is that since the element $\ce{B}$ is in HCP, the number of octahedral and tetrahedral voids will be 6 and 12, respectively, which means that $2/3 \times 12 = 8$ tetrahedral voids occupied by $\ce{A}$.

Since $\ce{B}$ is in HCP lattice, there must be 12 voids, i.e. the ratio will be $8 : 12 \implies 2 : 3$, so $\ce{A2B3}$ must be the answer.

But according to NCERT textbook (example 1.2) the answer is $\ce{A4B3}$.