Sets and Relations – Relations – JEE Main 24 Jan 2026 Shift 1

Solution:

We need to find the number of pairs $((a,b), (c,d))$ where $a,b,c,d \in \{1,2,3,4\}$ such that:
$$2a + 3b = 3c + 4d$$
Let $LHS = 2a + 3b$ and $RHS = 3c + 4d$.
Since $a,b,c,d \in \{1,2,3,4\}$, let’s analyze the possible values of LHS and RHS.

**Possible values of RHS ($3c+4d$):**
Min: $3(1)+4(1)=7$
Max: $3(4)+4(4)=28$

Let’s list valid combinations for RHS and check if they can be formed by LHS.

Case 1: $c=1$
$d=1 \Rightarrow RHS=7$. $2a+3b=7 \Rightarrow (2,1)$. (1 solution)
$d=2 \Rightarrow RHS=11$. $2a+3b=11 \Rightarrow (1,3), (4,1)$. (2 solutions)
$d=3 \Rightarrow RHS=15$. $2a+3b=15 \Rightarrow (3,3)$. (1 solution)
$d=4 \Rightarrow RHS=19$. $2a+3b=19$. Max possible LHS is $2(4)+3(4)=20$. Try $a=4, 3b=11$ (No). (0 solutions)

Case 2: $c=2$
$d=1 \Rightarrow RHS=10$. $2a+3b=10 \Rightarrow (2,2)$. (1 solution)
$d=2 \Rightarrow RHS=14$. $2a+3b=14 \Rightarrow (1,4), (4,2)$. (2 solutions)
$d=3 \Rightarrow RHS=18$. $2a+3b=18 \Rightarrow (3,4)$. (1 solution)
$d=4 \Rightarrow RHS=22 > 20$. (0 solutions)

Case 3: $c=3$
$d=1 \Rightarrow RHS=13$. $2a+3b=13 \Rightarrow (2,3)$. (1 solution)
$d=2 \Rightarrow RHS=17$. $2a+3b=17 \Rightarrow (1,5)$ (No), $(4,3)$. (1 solution)
$d=3 \Rightarrow RHS=21 > 20$. (0 solutions)

Case 4: $c=4$
$d=1 \Rightarrow RHS=16$. $2a+3b=16 \Rightarrow (2,4)$. (1 solution)
$d=2 \Rightarrow RHS=20$. $2a+3b=20 \Rightarrow (4,4)$. (1 solution)

Total Elements:
For $c=1, d=1$ (RHS=7): 1 pair
For $c=1, d=2$ (RHS=11): 2 pairs
For $c=1, d=3$ (RHS=15): 1 pair
For $c=2, d=1$ (RHS=10): 1 pair
For $c=2, d=2$ (RHS=14): 2 pairs
For $c=2, d=3$ (RHS=18): 1 pair
For $c=3, d=1$ (RHS=13): 1 pair
For $c=3, d=2$ (RHS=17): 1 pair
For $c=4, d=1$ (RHS=16): 1 pair
For $c=4, d=2$ (RHS=20): 1 pair

Sum $= 1 + 2 + 1 + 1 + 2 + 1 + 1 + 1 + 1 + 1 = 12$.

Ans. (3)