Maths > Functions > 2.0 Relation

  Functions
    1.0 Definitions
    2.0 Relation
    3.0 Types of Relation
    4.0 Functions
    5.0 Standard Real Functions and their Graphical Representation
    6.0 Operations on Real Functions
    7.0 Types of Functions
    8.0 Composition of a Function
    9.0 Inverse of a Function

2.1 Total Number of Relations
Let $A$ and $B$ be two non-empty finite sets consisting of $m$ and $n$ elements respectively. Then $A \times B$ consists of $mn$ ordered pairs. So, total number of subsets of $A \times B$ is ${2^{mn}}$.

And, since each subset of $A \times B$ defines a relation from $A$ to $B$, so total number of relations from $A$ to $B$ is ${2^{mn}}$.

Among these, the void relation $(\emptyset )$ and the universal relation $A \times B$ are trivial relations from $A$ to $B$.


Example 10:.Find the total number of relations possible in each case, where relation is from $A$ to $B$

i. $\begin{equation} \begin{aligned} A = \{ 2,3,5,7\} \\ B = \{ 0\} \\\end{aligned} \end{equation} $ii. $\begin{equation} \begin{aligned} A = \emptyset \\ B = \{ 1,2,3,4\} \\\end{aligned} \end{equation} $iii. $\begin{equation} \begin{aligned} A = \{ x:x \in N\} \\ B = \{ a,e,i,o,u\} \\\end{aligned} \end{equation} $
iv. $\begin{equation} \begin{aligned} A = \{ {a_1},{a_2},{a_3},.......,{a_{n - 2}},{a_{n - 1}}\} \\ B = \{ {b_1},{b_2},{b_3},.........,{b_{{n^2} - 2}},{b_{{n^2} - 1}}\} \\\end{aligned} \end{equation} $ v. $\begin{equation} \begin{aligned} A = \{ x:x\;is\;a\;letter\;in\;English\;alphabet\} \\ B = \{ x:x \le 10\;and\;x \in N\} \\\end{aligned} \end{equation} $


Solution:

RelationTotal number of relations
i. $\begin{equation} \begin{aligned} A = \{ 2,3,5,7\} \\ B = \{ 0\} \\\end{aligned} \end{equation} $Number of elements in $A$, i.e. $n(A) = 4 = m$
Number of elements in $B$, i.e. $n(B) = 1 = n$
Total number of relations possible is ${2^{mn}} = {2^{4 \times 1}} = {2^4} = 16$
ii. $\begin{equation} \begin{aligned} A = \emptyset \\ B = \{ 1,2,3,4\} \\\end{aligned} \end{equation} $Number of elements in $A$, i.e. $n(A) = 0 = m$
Number of elements in $B$, i.e. $n(B) = 4 = n$
Total number of relations possible is ${2^{mn}} = {2^{0 \times 4}} = {2^0} = 1$
iii. $\begin{equation} \begin{aligned} A = \{ x:x \in N\} \\ B = \{ a,e,i,o,u\} \\\end{aligned} \end{equation} $Number of elements in $A$ is infinite.
Number of elements in $B$, i.e. $n(B) = 5 = n$
Since number of elements in set $A$ is infinite, total number of relations cannot be defined.
iv. $\begin{equation} \begin{aligned} A = \{ {a_1},{a_2},{a_3},.......,{a_{n - 2}},{a_{n - 1}}\} \\ B = \{ {b_1},{b_2},{b_3},.........,{b_{{n^2} - 2}},{b_{{n^2} - 1}}\} \\\end{aligned} \end{equation} $Number of elements in $A$, i.e. $n(A) = n-1 = k$
Number of elements in $B$, i.e. $n(B) = {n^2} - 1 = m$
Total number of relations possible is ${2^{km}} = {2^{(n - 1)({n^2} - 1)}} = {2^{{n^3} - {n^2} - n + 1}},\;n \ge 1$
v. $\begin{equation} \begin{aligned} A = \{ x:x\;is\;a\;letter\;in\;English\;alphabet\} \\ B = \{ x:x \le 10\;and\;x \in N\} \\\end{aligned} \end{equation} $Number of elements in $A$, i.e. $n(A) = 26 = m$
Number of elements in $B$, i.e. $n(B) = 10 = n$
Total number of relations possible is ${2^{mn}} = {2^{26 \times 10}} = {2^{260}}$

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