Question

A set containing n elements. A subset P of A is chosen. The set A is reconstructed replacing the elements of P. A subset Q of A is again chosen. The number of ways of choosing P and Q such that $P\cap Q=\phi$ is:
(a) ${{2}^{2n}}{{-}^{2n}}{{C}_{n}}$
(b) ${{2}^{n}}$
(c) ${{2}^{n}}-1$
(d) ${{3}^{n}}$

Answer 1

Hint: To solve the above question, you need to apply the definition of the subsets and also the knowledge related selections and combinations. Focus on the point that each element in the set A has 3 options, i.e., either they will be present in the subset P else they will be present in set Q else they won’t be present in any of the subsets.

Complete step-by-step answer:
Before starting with the solution, let us discuss different symbols and operations related to the question.
Subset: A set A is said to be the subset of set B, if all the terms of A are present in set B, i.e., set A is contained in set B. This can be represented as: $A\subset B$ .
Now let us start with the solution to the question. We know that there are n elements. Now let us interpret the question as: there are 3 boxes, one box contains those elements which are present in the subset P, second contains elements of Q and the third contains those elements which are not present in the subsets. So, each element has 3 options, one to go in the box with elements present in the subset P, to go to the other box with elements of set Q and to go to the third box. So, we can say that the total number of possibilities of choosing P and Q such that $P\cap Q=\phi$ is ${{3}^{n}}$ .
Hence, the answer to the above question is option (d).

Note: Be careful about what is asked in the question. As the number of non-empty subsets, subsets and proper subsets are three different quantities and have different answers for a given set. Also, to use the above method, it is a necessary condition that $P\cap Q=\phi$ , else the number of possibilities for each element would increase, as one more possibility that the element is present in both P and Q would be added.