 Solution

Use fractional coordinates relative to conventional cubic vectors $(a\hat{x},a\hat{y},a\hat{z})$. A body-centered cubic lattice has representative pointsso its conventional cell contains two lattice points: eight shared corners contribute one and the interior point contributes one. A face-centered cubic lattice hasso its conventional cell contains four: one from the shared corners and three from six half-shared faces.

For $N$ atoms at fractional positions $(x_j,y_j,z_j)$, the geometric structure factor and the atomic-factor-weighted amplitude are respectivelyFor identical atoms $F=fG$. Applying the conventional-cell lattice factors givesandThese are lattice selection rules; an additional nontrivial crystal basis structure factor can extinguish further allowed reflections.

Of the eight equally represented parity triples, four satisfy the BCC condition and two the FCC condition. Thus the fractions among integer index triples are $1/2$ and $1/4$, respectively. This reflects the two and four equivalent lattice points in the conventional cells: their reciprocal lattices occupy corresponding subsets of the conventional reciprocal grids. In primitive reciprocal coordinates every integer triple denotes a reciprocal-lattice vector; lattice-centering absences are then built into the grid rather than an extra rule. A multi-atom chemical basis can still cause genuine additional absences. If permutations means merely reordering one fixed triple, its allowedness never changes: the quoted fractions concern all index triples, not reordering a chosen $(h,k,l)$.