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Here the [Cr(NH3) 4Cl2]^+ has a positive charge. 
NH3 has zero charge.
Cl has one -ve charge.
Let the oxidation number of Cr be x. 
So simple mathematical equation for charge on compound will be -
x + 4*0 + 2*(-1) = +1 
Solving we get 
x = +3

So, Oxidation number of Cr in [Cr(NH3) 4Cl2]^+ is +3

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