How to find oxidation state of Co and Cr in [Co(NH₃)₄(NO₂)₂][Cr(NH₃)₃(NO₂)₃]? Which of the following options represents the oxidation state of $\ce{Co}$ and $\ce{Cr}$ in the given complex?
$$\ce{[Co(NH3)4(NO2)2][Cr(NH3)3(NO2)3]}$$
(A) $2, 3$
(B) $3, 2$
(C) $3, 3$
(D) $2, 2$

I am unable to judge the net charge on complex when it breaks into ions and therefore I am having trouble with this question. I assumed that some unknown charge with opposite signs must be on the complex ions.
Conventionally, first complex ion is positively charged and the latter is negatively charged. Suppose thess charges are $p+$ and $q-$, respectively. Since two ion ratio is 1:1, $|p+|= |q-|$.

Suppose oxidation number of $\ce{Co}$ is $n$ and that of $\ce{Cr}$ is $m$. Hence, for cation complex (Note that the net charges of $\ce{NH3}$ and $\ce{NO2-}$ is $0$ and $-1$, respectively):
  $$n+0\times4+(-1)\times2=+p = n-2$$
and for anion complex:
 $$m+0\times3+(-1)\times3=-q=m-3$$
Since, $|p+|= |q-|$,
$$m-3= +p \; \text{and  }  n-2=-p$$
Or, $m+n=5$. 
The only answer agree with this is (A) or (B):

If (A) is correct, then, $\ce{Co^2+}$ and $\ce{Cr^3+}$. Substitute these values in original two equation: $+2+0\times4+(-1)\times2=+0 $ and $+3+0\times4+(-1)\times3=+0 $. This can't be.

If (B) is correct, then, $\ce{Co^3+}$ and $\ce{Cr^2+}$. Substitute these values in original two equation: $+3+0\times4+(-1)\times2=+1 $ and $+2+0\times4+(-1)\times3=-1 $. This is correct.

Therefore, your answer is (B), and the oxidation numbers of $\ce{Co}$ and $\ce{Cr}$ ared $+3$ and $+2$, respectively.