For fcc, number of X atoms = 4/unit cell

Number of Tetrahedral Voids = Z = 8

Number of Octahedral Voids = Y = 4

Number of atoms removed along one body diagonal = 2X (corner) and 2Z (TVs) and 1 Y (OV at body centre)

$\therefore$    Number of X atoms left $=4-\left(2\times \frac{1}{8}\right)=\frac{15}{4}$

Number of Y atom left = 4 - (1 × 1) = 3

Number of Z atom left = 8 - (2 × 1) = 6

The simplest formula $={\text{X}}_{\frac{15}{4}}{\text{Y}}_3{\text{Z}}_6\Rightarrow {\text{X}}_{15}{\text{Y}}_{12}{\text{Z}}_{24}$
                                                     $\Rightarrow {\text{X}}_5{\text{Y}}_4{\text{Z}}_8$