Question

Consider a set S = {1, 2, 3, ..., 6}. The total
number of ordered pairs (A, B), where A and
B are different subsets of S such that
A B = {1,2}, is ...
(A) 78
(B) 79
(C) 80
(D) none of these​

Answer 1

Answer:

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Answer 2

Answer:

If A and B are disjoint sets, then A∩B=ϕ.

We have to choose A and B such that they are disjoint subsets of S={1,2,3,4}.  

They have to be unordered pair too.

Each element in S can be an element of A or of B or of neither subsets.  

For each element, there are three possibilities.  

Hence for four elements there are 3  

4

 possibilities.  

Now this contains ordered pairs also except for the case where both A and B are null sets (This appears only once).  

Hence total number of ordered pairs of subsets =3  

4

+1=82

∴ total number of unordered pairs=  

2

82

​  

=41

Step-by-step explanation:

hope it helps