Question

The number of functions $f$ from $\left\{1,2, 3, \ldots, 20\right\}$ onto $\left\{1,2, 3, \ldots, 20\right\}$ 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

Domain and codomain $=\left\{1,2, 3, \ldots20 \right\}$
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
$=^{6}C_{5}\times5! = 6!$
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 $=15! . 6!$
$\Rightarrow a \times b = 15 \times 6 = 90$