JEE Main · 2022 · 28 Jun 2022 · Shift 2 · Permutations and Combinations

Because $f$ is symmetric and $f(a,b)\ge a$, every value satisfies $f(a,b)\ge \max (a,b)$, so $f$ is fixed by its values on the ten unordered pairs. Any pair whose larger element is $4$ must map to $4$, so $\{1,4\},\{2,4\},\{3,4\},\{4,4\}$ are forced to $4$ and the value $4$ is automatically present. The value $1$ can come only from $\{1,1\}$, so onto-ness forces $f(1,1)=1$. That leaves $\{1,2\},\{2,2\}$ free over $\{2,3,4\}$ and $\{1,3\},\{2,3\},\{3,3\}$ free over $\{3,4\}$, giving $3^2\cdot 2^3=72$ assignments. Requiring both $2$ and $3$ to appear, we subtract the $2^2\cdot 2^3=32$ assignments missing $2$ and the $2^2\cdot 1^3=4$ missing $3$, then add back the single assignment missing both: $72-32-4+1=37$.