A=1/2b(c+d) solve for c
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The inclusion/exclusion principle states that
That is, the union has as many members as the sum of the number of members of the individual sets, minus the number of elements contained in both sets (to avoid double-counting).
Therefore,
will have the most elements when the sets 
and 
are disjoint, i.e. 
, which would mean the most we can can in this case would be
(Note that
denotes the cardinality of the set .)
That is, the union has as many members as the sum of the number of members of the individual sets, minus the number of elements contained in both sets (to avoid double-counting).
Therefore,
(Note that