The total number of stereoisomers of the octahedral ion $[\mathrm{Co}(\mathrm{en})(\mathrm{NH}{}_3){}_2\mathrm{Cl}{}_2]{}^{+}$ (en = symmetrical bidentate) is

This is $\mathrm{M}(\mathrm{AA})\mathrm{b}{}_2\mathrm{c}{}_2$. With the two $\mathrm{NH}{}_3$ trans, the two $\mathrm{Cl}{}^{-}$ must also be trans (1 achiral). With the $\mathrm{NH}{}_3$ pair cis there are two arrangements: $\mathrm{Cl}{}^{-}$ trans to en N, which is achiral, or the two $\mathrm{Cl}{}^{-}$ cis, which is chiral (2 enantiomers). Total $=1+1+2=4$.