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NCERT Exemplar · Q24

Q.Let n(A)=mn(A) = m, and n(B)=nn(B) = n. Then the total number of non-empty relations that can be defined from AA to BB is
(A) mnm^n
(B) nm−1n^m - 1
(C) mn−1mn - 1
(D) 2mn−12^{mn} - 1

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A relation from set AA to set BB is any subset of their Cartesian product A×BA \times B. If n(A)=mn(A)=m and n(B)=nn(B)=n, then n(A×B)=mnn(A \times B) = mn, leading to 2mn2^{mn} total relations. Excluding the single empty relation, the number of non-empty relations is 2mn−1\boxed{2^{mn} - 1}.

Let's break down the concept of relations and how to count them. The core idea is that a relation between two sets is fundamentally defined by which pairs of elements are "related." This naturally leads us to consider the Cartesian product of the sets.

A relation RR from set AA to set BB is simply a collection of ordered pairs (a,b)(a, b) where a∈Aa \in A and b∈Bb \in B. For example, if A={1,2}A = \{1, 2\} and B={x,y}B = \{x, y\}, a relation could be R={(1,x),(2,y)}R = \{(1, x), (2, y)\}. Another relation could be R′={(1,y)}R' = \{(1, y)\}. Notice that these relations are nothing more than subsets of all possible ordered pairs you can form between elements of AA and BB. This set of all possible ordered pairs is called the Cartesian product, A×BA \times B.

1. Determine the size of the Cartesian Product A×BA \times B

The Cartesian product A×BA \times B is the set of all possible ordered pairs (a,b)(a, b) where a∈Aa \in A and b∈Bb \in B.

Given:

  • The number of elements in set AA is n(A)=mn(A) = m.
  • The number of elements in set BB is n(B)=nn(B) = n.

To form an ordered pair (a,b)(a, b), we choose one element from AA (there are mm choices) and one element from BB (there are nn choices). By the fundamental principle of counting, the total number of such distinct ordered pairs is the product of the number of choices for each position.

Therefore, the number of elements in the Cartesian product A×BA \times B is:

n(A×B)=n(A)×n(B)=m×n=mnn(A \times B) = n(A) \times n(B) = m \times n = mn.

For example, if A={1,2}A = \{1, 2\} (m=2m=2) and B={x,y,z}B = \{x, y, z\} (n=3n=3), then A×B={(1,x),(1,y),(1,z),(2,x),(2,y),(2,z)}A \times B = \{(1, x), (1, y), (1, z), (2, x), (2, y), (2, z)\}. Here, n(A×B)=2×3=6n(A \times B) = 2 \times 3 = 6.

2. Calculate the total number of possible relations

A relation from AA to BB is defined as any subset of A×BA \times B.

Let S=A×BS = A \times B. We found that n(S)=mnn(S) = mn.

If a set has kk elements, then the total number of its subsets (also known as its power set) is 2k2^k.

Since a relation is a subset of A×BA \times B, and A×BA \times B has mnmn elements, the total number of possible relations that can be defined from AA to BB is 2mn2^{mn}.

This count includes all possible subsets, from the empty set to the set A×BA \times B itself. …

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