Number of optically active isomers of [Co(NH3)2Cl2(NO)(OH)]? My Attempt:

Since Compound is of the form M[a2b2cd] 

I got three optically active isomers of the form M[(ab)(ab)(cd)] , M[(ab)(ac)(bd)] , M[(ab)(bc)(ad)]
and since NO is a bidentate ligand, total number of optically active isomers should be multiplied by 2 due to linkage isomerism.

Hence i was getting the answer as 2*3=6 but answer given was 4. 

Can you please help me with this?
Try drawing out [M(ab)(ab)(cd)]: you'll see that it's actually optically inactive!



So, there are 2 optically active isomers, which along with their respective enantiomers adds up to 4 (M[(ab)(ac)(bd)] , M[(ab)(bc)(ad)],  as you've mentioned).