{0}

{5}

All the possible positions of the value {1} in {4} different {2}s are in the same {4} {3}s. Because the number of {2}s matches the number of {3}s, and because each {2} must contain the value {1}; each {3} will have a {1} in one of the intersecting cells.

Hence other possible positions of the value {1} that are in the {3}s but not in one of the {2}s can be removed.

Therefore we can remove {6}.