Search any question & find its solution
Question:
Answered & Verified by Expert
If $A=\left\{x \in z^{+}: x < 10\right.$ and $x$ is a multiple of 3 or $4\}$, where $z^{+}$is the set of positive integers, then the total number of symmetric relations on $A$ is
Options:
Solution:
1033 Upvotes
Verified Answer
The correct answer is:
$2^{15}$
$2^{15}$
A relation on a set A is said to be symmetric iff $(a, b) \in A \Rightarrow(b, a) \in A, \forall a, b \in A$
Here $A=\{3,4,6,8,9\}$
Number of order pairs of $A \times A=5 \times 5=25$
Divide 25 order pairs of $A \times A$ in 3 parts as follows :
$$
\begin{aligned}
& \text { Part }-A:(3,3),(4,4),(6,6),(8,8),(9,9) \\
& \text { Part }-B:(3,4),(3,6),(3,8),(3,9),(4,6), \\
& (4,8),(4,9),(6,8),(6,9),(8,9) \\
& \text { Part-C:(4,3),(6,3),(8,3),(9,3),(6,4),(8,4), } \\
& (9,4),(8,6),(9,6),(9,8)
\end{aligned}
$$
In part $-A$, both components of each order pair are same.
In part $-B$, both components are different but not two such order pairs are present in which first component of one order pair is the second component of another order pair and vice-versa.
In part $-C$, only reverse of the order pairs of part $-B$ are present i.e., if $(a, b)$ is present in part $-B$, then (b, a) will be present in part $-C$ For example $(3,4)$ is present in part $-B$ and $(4,3)$ present in part $-C$.
Number of order pair in $A, B$ and $C$ are 5, 10 and 10 respectively.
In any symmetric relation on set $A$, if any order pair of part $-B$ is present then its reverse order pair of part $-C$ will must be also present.
Hence number of symmetric relation on set $A$ is equal to the number of all relations on a set $D$, which contains all the order pairs of part $-A$ and part $-B$.
Now $n(D)=n(A)+n(B)=5+10=15$
Hence number of all relations on set $D=(2)^{15}$
$\Rightarrow$ Number of symmetric relations on set $D=(2)^{15}$
Here $A=\{3,4,6,8,9\}$
Number of order pairs of $A \times A=5 \times 5=25$
Divide 25 order pairs of $A \times A$ in 3 parts as follows :
$$
\begin{aligned}
& \text { Part }-A:(3,3),(4,4),(6,6),(8,8),(9,9) \\
& \text { Part }-B:(3,4),(3,6),(3,8),(3,9),(4,6), \\
& (4,8),(4,9),(6,8),(6,9),(8,9) \\
& \text { Part-C:(4,3),(6,3),(8,3),(9,3),(6,4),(8,4), } \\
& (9,4),(8,6),(9,6),(9,8)
\end{aligned}
$$
In part $-A$, both components of each order pair are same.
In part $-B$, both components are different but not two such order pairs are present in which first component of one order pair is the second component of another order pair and vice-versa.
In part $-C$, only reverse of the order pairs of part $-B$ are present i.e., if $(a, b)$ is present in part $-B$, then (b, a) will be present in part $-C$ For example $(3,4)$ is present in part $-B$ and $(4,3)$ present in part $-C$.
Number of order pair in $A, B$ and $C$ are 5, 10 and 10 respectively.
In any symmetric relation on set $A$, if any order pair of part $-B$ is present then its reverse order pair of part $-C$ will must be also present.
Hence number of symmetric relation on set $A$ is equal to the number of all relations on a set $D$, which contains all the order pairs of part $-A$ and part $-B$.
Now $n(D)=n(A)+n(B)=5+10=15$
Hence number of all relations on set $D=(2)^{15}$
$\Rightarrow$ Number of symmetric relations on set $D=(2)^{15}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.