Here X and Y are representing subsets such that X ∩ Y = Φ meaning that X and Y are disjoint subsets..So for finding number of such pairs which are disjoint to each other and let the size of the set = n.
Let we have number of elements in X = r..
So number of ways of selecting elements in X such that it contains 'r' elements = nCr
So we have number of elements left for inclusion in Y = n-r..
Now each of these 'n-r' elements can be either selected or rejected , hence 2 choices for each element..
Hence total number of possibilities of Y such that it is disjoint to X = 2 . 2 . ......(n-r) times
= 2n-r
Hence number of pairs (X,Y) such that X contains 'r' elements = nCr * 2n-r
Now 'r' varies from 0 to n..
Hence total number of pairs (X,Y) = Σ nCr * 2n-r
= (1 + 2)n
= 3n
But (X,Y) is same as (Y,X) , hence double counting is to be avoided..
Hence number of such pairs = [ ( 3n - 1 ) / 2 ] + 1
Substituting n = 6 , we get number of pairs = 728 / 2 + 1
= 365