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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.

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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 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.

I think your confusion comes about from symmetry because some of the combinations you calculated can be rotated into others.

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