in recent years many questions have come in competitive exams which ask for the number of isomers (all types including stereo and geometric), so how should we make and draw these isomers on the spot particularly if the question is of integer answer type, where even counting one  extra or one less is equal to -1 marks.

For calculating stereoisomers ( optical and meso ) following scheme should be followed :

a) When the molecule is unsymmetrical and contains 'n' stereogenic centres or chiral carbon atoms :

Number of optical isomers, i.e., d and l = 2ⁿ ; Number of meso forms = zero

Total number of stereoisomers = 2ⁿ+ 0 =2ⁿ

b) When the molecule is symmetrical and has even number of stereogenic centres or chiral carbon atoms :

Number of optical isomers = 2^(n-1)  ; Number of meso forms = 2^{(n/2 -1)}

Total number of stereoisomers = 2^(n-1) + 2^{(n/2 -1)}

c)When the molecule is symmetrical and has odd number of stereogenic centres :

Number of optical isomers = 2^(n-1) - 2^{(n/2 -1/2)} ; Number of meso forms = 2^{(n/2 -1/2)}

Total number of stereoisomers = 2^(n-1)

For geometrical isomers, for every double bond with different groups we have two isomers cis and trans.

Structural isomers, you have to draw and count.

Practice questions on isomerism and you will find its not that difficult, rather it is more interesting .