# The number of geometrical isomers of complex of type [Ma3b2c]

The question is to find out the number of geometrical isomers of the complex $\ce{[Ma_3b_2c]}$.

I applied simple combinations to get the number of isomers as $\Large\frac{6!}{3!2!}$ which is 60 in total. However my book states that only 3 geometrical isomers are possible. I could not see why simple combinations is not giving correct answer.

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 3 $\ce{a}$'s. So that's one isomer.
If $\ce{a}$ is trans to $\ce{c}$, then the remaining 2 $\ce{a}$'s can be cis or trans to each other, i.e., all 3 $\ce{a}$'s can be facial or meridional, respectively. That's two more isomers. That's it.