Question

Let S=\left\\{ 1,2,3,4 \right\\}. The total number of unordered pairs of disjoint subsets of S is equal to,
A. 25
B. 34
C. 42
D. 41

Answer 1

We will first start by using the fact that if A and B are two disjoint sets then $A\cap B=\phi$. Then we will find the different pair of subsets of S by using the fundamental principle of counting. We will first take the sets of all single elements of the S then we will take the subsets having two elements each and further we will take the subsets having the three and four elements each. Then will use the fact that if A and B are disjoint subsets then $A\cap B=\phi$ .Using this we will find the all unordered pairs of disjoint subsets.

Complete step-by-step answer:
Now, we have S=\left\\{ 1,2,3,4 \right\\}, we know that number of subset of a set having n elements is ${{2}^{n}}$ and therefore, the number of subset of S is ${{2}^{4}}$ i.e. 16.
Now, we have among 16 subsets 1 with zero element, 4 with one element, 6 with two elements, 4 with three elements and 1 with four elements because the ways of selecting r element from a set of n elements is ${}^{n}{{C}_{r}}$.
We will consider the pairs of elements starting with the empty set and ending with the universal set.
CASE – I: The empty set
We know that the empty set is disjoint with all other sets and with itself. Therefore, we get 16 pairs of disjoint subsets.
CASE – II: Singleton set
We have already considered the relation with an empty set.
So, among the sets with one element we have 6 pairs of disjoint sets.
Each singleton set is disjoint with 3 sets having two elements. Since, we have 4 singleton sets, therefore, this gives 12 pairs of disjoint sets.
Each singleton set is disjoint with 1 set having three elements. So, this gives us a total of 4 pairs of disjoint sets.
None of the sets with one element is disjoint with the set having four elements.
CASE – III: Set with two elements
Since, we have already considered the relation with the empty set and the singleton sets.
There are 3 pairs of disjoint sets among the sets having two elements.
CASE – IV: Set with three elements
None of the sets with three elements is not disjoint with sets having three or four elements as we can see that there are only four elements in the set. So, if the subset has three or more elements then one element has to be common.
CASE – V: Set with four elements
The set that is the universal set is not disjoint with itself.
So, now we have two pairs of disjoint subsets from all the cases as,
$16+6+12+4+3=41$
Hence, the correct option is (D).

Note: It is important to note that we have splitted the question into different cases and have counted unordered pairs of subsets for each case. Also, it is important to note that the number of pair of disjoint subset with one element is 6 as the ways of choosing 2 elements among 4 is ${}^{4}{{C}_{2}}$
$=\dfrac{4!}{2!\times 2!}=3\times 2=6$.