A crystal made up of particles X, Y, and Z. X forms fcc packing. Y...

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)

$∴$ Number of X atoms left $= 4 - \left(2 \times \dfrac{1}{8}\right) = \dfrac{1 5}{4}$

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

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

The simplest formula $= \text{X} \dfrac{1 5}{4} {Y}_{3} {Z}_{6} ⇒ {X}_{1 5} {Y}_{1 2} {Z}_{2 4}$
& & & & & & & & & & & & & & & & & & & & & & & & $⇒ {X}_{5} {Y}_{4} {Z}_{8}$