JEE Main 2022 · 24 Jun · Shift 1, Mathematics question 21

Rewrite. $f(b)=2f(a)+3f(c)+f(d)\le10$, with all four values distinct in $\{0,\dots,10\}$.

Count by $f(c)$ (and then $f(a)$; $f(d)$ is free subject to distinctness and $f(b)\le10$):

• $f(c)=0$: $f(a)=1,2,3,4$ give 7, 5, 3, 2 choices of $f(d)$: 17.
• $f(c)=1$: $f(a)=0,2,3$ give 6, 2, 1: 9.
• $f(c)=2$: $f(a)=0,1$ give 3, 1: 4.
• $f(c)=3$: $f(a)=0$ gives 1: 1.

(For example, $f(c)=0,f(a)=1$: $f(b)=2+f(d)\le10$ with $f(d)\notin\{0,1\}$ and $f(b)\ne f(d)$ leaves $f(d)=2,\dots,8$.)

Total $17+9+4+1=31$.

Answer: 31