Question

The number of functions $f$ from $\left\{1,2,3,\dots ,20\right\}$ onto $\left\{1,2,3,\dots ,20\right\}$ such that $f(k)$ is a multiple of 3, whenever $k$ is a multiple of 4 is of the form $a!\times b!$ then $a\times b$ is

Answer 1

Answer: 90

Domain and codomain $=\left\{1,2,3,\dots 20\right\}$
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, whenever $k$ 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 $=15!.6!$
$\Rightarrow a\times b=15\times 6=90$