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the number of subsets from k elements is 2^k
here if we consider a and c together as one element (since they are always present together), won't the number of subsets: 2^3 = 8?
now a&c will always need to be present, therefore the only remaining subsets: 2^2 = 4

is above understanding correct?
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Yes correct Number of subsets for a set is given by 2^k, in your case you consider 2^3

but this will also include following subset: {}, {b}, {d},{a,c},{a,c,d},{a,c,b},{b,d},{a,c,b,d} = 8 sets

The question asks for sets with both a&c, hence the answer is {a,c},{a,c,d},{a,c,b},{a,c,b,d} = 4 sets