Suppose for example the compounds $\ce{M(AA)3}$ and $\ce{M(AA)3bc}$, with $\ce{M}$, the central metal atom and $\ce{AA}$, a bidentate ligand.

What is the total number of stereo-isomers of both?

I was taught that for the compound $\ce{M(AA)3}$ the number of stereo isomers is 2; there are no geometric isomers and one is optically active. For $\ce{M(AA)3bc}$, the total of stereo isomers is 3.


$$ \text{number of stereo isomers} = \text{number of geometric isomers} +\text{number of optical isomers} $$

then why the total of stereo isomers is 2 for $\ce{M(AA)3}$ if only one structure is optically active, while the above formula stands good for $\ce{M(AA)3bc}$?


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