Shortcuts for finding geometrical isomers

Our teacher provides us extra questions for pratice, so one of the questions was :-



> 
> Find number of products if following are reacted with excess $\ce{NH2OH}$ (hydroxyl amine):
> 
> 
> Cyclohex -   
> 
> (i) 1,4-dione   
> 
> (ii) 1,3-dione   
> 
> (iii) 1,2-dione   
> 
> (iv) 1,2,3 trione   
> 
> (v) 1,2,4-trione   
> 
> (vi) 1,3,5-trione and   
> 
> (vii) cyclobut-1,2,3,4-tetraone
> 
> 
> 


To solve this, our teacher told us replace each $\ce{=O}$ group by 1 $\ce{-CH3}$ and 1 $\ce{-H}$ , then find the number of stereoisomers (including enantiomers) of the resultant compound, which is the answer.(if you actually count it's tough to visualize the lone pair and $\ce{-OH}$ bond and it gets long).


I have cross-checked , the answers pretty much match. So my doubt is, is this trick always right and if yes then please prove it. I would also be content if you can provide any links, if the question has already been answered.




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I had another doubt, about another trick my teacher told us about. Let's say there's a compound with $n$ carbon atoms forming a closed ring of all $n$ carbons. At each carbon, out of 2 $\ce{-H}$ , 1 $\ce{-H}$ is replaced by $\ce{-Cl}$ (chlorine). The trick if $n$ is odd ,the number of geometrical isomers of the compound is $2^{(\frac{n-1}{2})}$ and if $n$ is even then the number of geometrical isomers is $2^{(\frac{n}{2})}$. Please prove this also.


Thank you in advance.


