Considering both geometrical and linkage isomerism (octahedral, no optical isomers here), the total number of isomers of $[Co (NH X_{3}) X_{4} (NO X_{2}) Cl] X^{+}$ in which $NO X_{2} X^{-}$ may bind through N or O is

Answer: B

The complex is

Ma X_{4} bc

with 2 geometrical forms (cis and trans, neither chiral). Each form has 2 linkage options (nitro or nitrito). Total

= 2 \times 2 = 4