Question

If $A$ is the set of even natural numbers less than $8$ and $B$ is the set of prime numbers less than $7$, then the number of relations from $A$ to $B$ is
A. ${2^9}$
B. ${9^2}$
C. ${3^2}$
D. ${2^9} - 1$

Answer 1

Hint: In order to find the number of relations for the even numbers and prime numbers, we should know about the even numbers and the prime numbers. Even numbers which are completely divisible by $2$, gives the remainder zero. Prime numbers are the numbers which are divided by two numbers only, number $1$ and the number itself.Make the sets for the two and obtain a relation between them.

Complete step by step answer:
We are given two sets of numbers, the first is even numbers. Even numbers are numbers that are completely divisible by $2$. In this case we need even numbers that are less than $8$, so the numbers are $2,4,6$. When written in set, it becomes:
$A = \left\{ {2,4,6} \right\}$
Considering the number of elements in the set A to be m and counting the elements, we get:
$m = 3$ …..(1)

The set of prime numbers, which are less than $7$ are $2,3,5$. When written in set, it becomes:
$B = \left\{ {2,3,5} \right\}$
Considering the number of elements in the set B to be n and counting the elements, we get:
$n = 3$ …..(2)
From the theory of sets and relations, we know that the relation between two sets is equal to: $2$ to the power of product of the number of elements of two sets, and according to this:
The relation from the set A to B is:
${2^{mn}}$
Substituting the value of $m$ and $n$:
$ \Rightarrow {2^{3 \times 3}}$
$ \Rightarrow {2^9}$
Therefore, the Relation from A to B is ${2^9}$.

Hence, option A is correct.

Note:A relation is basically a set of pairs or ordered pairs formed from two sets. In an ordered pair, one element is from the first set and the second one from the other. The number of ordered pairs is decided by multiplying the number of elements in the first set with the number or elements from the second set.