JEE Main 2026 — Sets and Relations Question with Solution

Given the set $A=\{2,3,4,5,6\}$.

The relation $R$ is defined on $A\times A$ such that $(x,y)R(z,w)$ if and only if $x$ divides $z$ and $y\le w$.

The number of elements in $R$ is the number of ordered pairs $((x,y),(z,w))$ satisfying these conditions. Since the conditions for $(x,z)$ and $(y,w)$ are independent, the total number of elements in $R$ will be the product of the number of pairs $(x,z)$ satisfying $x$ divides $z$ and the number of pairs $(y,w)$ satisfying $y\le w$.

First, we find the number of pairs $(x,z)\in A\times A$ such that $x$ divides $z$:
For $x=2$, $z\in \{2,4,6\}$ ($3$ pairs)
For $x=3$, $z\in \{3,6\}$ ($2$ pairs)
For $x=4$, $z\in \{4\}$ ($1$ pair)
For $x=5$, $z\in \{5\}$ ($1$ pair)
For $x=6$, $z\in \{6\}$ ($1$ pair)
Total number of pairs $(x,z)=3+2+1+1+1=8$.

Next, we find the number of pairs $(y,w)\in A\times A$ such that $y\le w$:
For $y=2$, $w\in \{2,3,4,5,6\}$ ($5$ pairs)
For $y=3$, $w\in \{3,4,5,6\}$ ($4$ pairs)
For $y=4$, $w\in \{4,5,6\}$ ($3$ pairs)
For $y=5$, $w\in \{5,6\}$ ($2$ pairs)
For $y=6$, $w\in \{6\}$ ($1$ pair)
Total number of pairs $(y,w)=5+4+3+2+1=15$.

The total number of elements in the relation $R$ is the product of the number of these pairs:
Total elements = $8\times 15=120$.

Answer: $120$