JEE Main · 2022 · 24 Jun 2022 · Shift 1 · Permutations and Combinations

The condition $2f(a)-f(b)+3f(c)+f(d)=0$ rearranges to $f(b)=2f(a)+3f(c)+f(d)$, so once $f(a),f(c),f(d)$ are chosen the value of $f(b)$ is fixed; the four images must still be distinct and all must lie in $\{0,1,\dots ,10\}$, which forces $f(b)\le 10$. Because $f(c)$ is multiplied by $3$ it is convenient to count according to its value. When $f(c)=0$ the requirement $2f(a)+f(d)\le 10$ with $f(a),f(d)$ distinct and non-zero yields seventeen admissible choices; when $f(c)=1$ it drops to nine; when $f(c)=2$ to four; and when $f(c)=3$ only one choice survives, while $f(c)\ge 4$ already makes $3f(c)$ too large. In each surviving case the resulting $f(b)$ differs from the other three images, so every choice is a genuine one-one function, and adding the cases gives $17+9+4+1=31$ such functions.