Question

If we have a set as $S = \\{ 1,2,3\\}$ then the total number of unordered pairs of disjoint subsets of S is
$A)25$
$B)34$
$C)42$
$D)14$

Answer 1

First, we will need to know about the concept of the unordered pairs and disjoint sets.
The two or more sets are said to be disjoint sets if they have no elements in common, which means their intersection will only yield the empty set like $A \cap B \ne \phi$, while the unordered pairs are the sets having no particular relation between them and denoted as $\\{ a,b\\}$
Formula used:
We make use of the concept of the total number of unordered pairs of disjoint subsets of any set containing n-elements is $\dfrac{{{3^n} + 1}}{2}$

Complete step-by-step solution:
Since from the given that we have the set as $S = \\{ 1,2,3\\}$. Now we need to find the total number of unordered pairs of disjoint subsets of S.
Now we need to find out the total number of subsets of the given set S, such that the subsets are unordered pairs of disjoint subsets of S.
We know that the total number of the unordered pairs of disjoint subsets of any set containing n-elements is equal to $\dfrac{{{3^n} + 1}}{2}$
Also, from the given that we have $S = \\{ 1,2,3\\}$ and clearly if we see there are total $3$ elements in the set, which means $n = 3$ and substitute into the original formula we get $\dfrac{{{3^n} + 1}}{2} = \dfrac{{{3^3} + 1}}{2}$
Now ${3^3}$ is nothing but $3 \times 3 \times 3 = 81$ and thus we have $\dfrac{{{3^3} + 1}}{2} = \dfrac{{27 + 1}}{2}$
Further solving we get $\dfrac{{27 + 1}}{2} = \dfrac{{28}}{2} = 14$ (by division operation)
Hence, the total number of unordered pairs of disjoint subsets of S is $14$
Therefore, the option $D)14$ is correct.

Note: If we face these kinds of similar questions, we use the concept of the total number of unordered pairs of the disjoint subsets, which is in the set theory.
Intersection empty means disjoint and union of the two subsets given us the whole set if the two subsets containing every element of the universal sets.
Also, power set means ${2^n}$ where n is the number of elements in the given set.