JEE Main 2026 — Sets and Relations Question with Solution

Let $A=\{1,4,7\}$ and $B=\{2,3,8\}$. The relation $R$ consists of pairs $((a_1,b_1),(a_2,b_2))$ from $A\times B$ such that $(a_1+b_2)$ divides $(a_2+b_1)$.

Let $x=a_1+b_2$ and $y=a_2+b_1$. Note that $(a_1,b_2)$ and $(a_2,b_1)$ are both independent elements of $A\times B$. So choosing $((a_1,b_1),(a_2,b_2))$ is equivalent to independently choosing two elements of $A\times B$ with sums $x$ and $y$ such that $x\mid y$.

All possible sums $a+b$ with $a\in A$ and $b\in B$:
$1+2=3$, $1+3=4$, $1+8=9$
$4+2=6$, $4+3=7$, $4+8=12$
$7+2=9$, $7+3=10$, $7+8=15$

Frequencies of each sum:
$f(3)=1$, $f(4)=1$, $f(6)=1$, $f(7)=1$, $f(9)=2$, $f(10)=1$, $f(12)=1$, $f(15)=1$

For each possible divisor $x$, count the number of valid pairs $(x,y)$ with $x\mid y$, each contributing $f(x)\cdot f(y)$:

$x=3$: divides $3,6,9,12,15$
Ways $=1\cdot (1+1+2+1+1)=6$

$x=4$: divides $4,12$
Ways $=1\cdot (1+1)=2$

$x=6$: divides $6,12$
Ways $=1\cdot (1+1)=2$

$x=7$: divides $7$
Ways $=1\cdot 1=1$

$x=9$: divides $9$
Ways $=2\cdot 2=4$

$x=10$: divides $10$
Ways $=1\cdot 1=1$

$x=12$: divides $12$
Ways $=1\cdot 1=1$

$x=15$: divides $15$
Ways $=1\cdot 1=1$

Total number of elements in $R$:
$6+2+2+1+4+1+1+1=18$

Hence, the answer is $18$.