JEE Main 2012 — Sets and Relations Question with Solution

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 : $$\begin{array}{rl}& \text{ Part }-A:(3,3),(4,4),(6,6),(8,8),(9,9)\\ & \text{ Part }-B:(3,4),(3,6),(3,8),(3,9),(4,6),\\ & (4,8),(4,9),(6,8),(6,9),(8,9)\\ & \text{ Part-C:(4,3),(6,3),(8,3),(9,3),(6,4),(8,4), }\\ & (9,4),(8,6),(9,6),(9,8)\end{array}$$ 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 $D=(2{)}^{15}$