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Let S={1,2,3,4}.Total no of unordered pairs of disjoint subsets of S is equal to

1)25   2)34   3)42  4)41

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Let $X$ and $Y$ be two disjoint subsets of $S$. 

Now, for each of the 4 elements of $S$ we have 3 choices:

  1. Go to $X$ and not $Y$.
  2. Go to $Y$ and not $X$.
  3. Go to neither $X$ nor $Y$. 

Thus we get two disjoint subsets $X$ and $Y$ in $3^4 = 81$ ways. But

  • we need to count unordered pairs- $ (X = \{1,2\}, Y =  \{3,4\})$ and  $ (X = \{3,4\}, Y = \{1,2\})$ should not be counted separate.

We can see that every pair $X,Y$ have two orders except $(X = \emptyset, Y = \emptyset)$. So, to count the no. of unordered pairs we can do $$\frac{81 - 1}{2} + 1 = 41.$$

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