Let x be the total number of stereoisomers of the octahedral complex [Co(gly)3] (where gly^-=NH2CH2COO^- binds through N and O as an unsymmetrical…

For $\mathrm{[Co(gly)_3]}$, a chelate $\mathrm{M(AB)_3}$, the three N atoms may be arranged facially (fac, all three N mutually cis) or meridionally (mer). Each of fac and mer lacks a mirror plane and is chiral (a propeller-like $\Delta/\Lambda$ pair), giving $2\times2=4$ stereoisomers, so $x=4$. For the square planar complex of type $\mathrm{Mabcd}$, fix one ligand, say $\mathrm{NH_3}$, and place the three others opposite it: the trans partner of $\mathrm{NH_3}$ may be py, Cl or Br, giving $y=3$ geometrical isomers (square planar complexes of this type are not chiral because the molecular plane is a mirror plane). Both counts were confirmed by enumerating all labelled arrangements modulo the rotation group of the octahedron (24 rotations, with chelate pairs required to be cis), which gave 4, and modulo the symmetry operations of the square plane, which gave 3. Hence $x+y=4+3=7$.