The number of geometrical isomers of complex of type [Ma₃b₂c]
First, let's establish that we're all talking about octahedral complexes to eliminate any possible confusion.
Fix $\ce{c}$. You can have either $\ce{b}$ trans or $\ce{a}$ trans.
If $\ce{b}$ is trans to $\ce{c}$, there's only one distinct way to arrange the remaining $\ce{b}$ and three $\ce{a}$'s. So that's one isomer.
If $\ce{a}$ is trans to $\ce{c}$, then the remaining two $\ce{a}$'s can be cis or trans to each other, i.e., all three $\ce{a}$'s can be facial or meridional, respectively. That's two more isomers. That's it.
I think your confusion comes about from symmetry because some of the combinations you calculated can be rotated into others.