JEE Main 2019 — Functions Question with Solution

Domain and codomain $=\{1,2,3,\dots ,20\}$. There are five multiple of 4 as 4,8,12,16 and 20 . and there are 6 multiple of 3 as 3,6,9,12,15,18 . Since, when ever $k$ is multiple of 4 then $f(k)$ is multiple of 3 then total number of arrangement $={}^6{c}_5\times 5!=6!$ Remaining 15 elements can be arranged in $15!$ ways. Since, for every input, there is an output $\Rightarrow$ function $f(k)$ in onto $\therefore$ Total number of arrangement $=15!.6!$