In a solid 'AB' having the NaCI structure, ‘A’ atoms occupy the corners of the cubic unit cell. If all the face-centered atoms along one of the axes are removed, then the resultant stoichiometry of the solid is

There were 6A atoms on the face-centres removing face centred atoms along one of the axes means removal of 2A atoms. 

Now, number of A atoms per unit cell

$=8\times \frac{1}{8}+4\times \frac{1}{2}=3$

      (corners) (face-centred)

Number of B atoms per unit cell

$=12\times \frac{1}{4}+1=4$

     (edge centred)  (body centred) 

Hence the resultant stoichiometry is A₃B₄