Question

The number of functions $f$ from $\{1,2,3,....,20\}$ onto $\{1,2,3,...,20\}$ such that $f\left(k\right)$ is a multiple of $3$, whenever $k$ is a multiple of $4$, is :

• A. $5!\times 6!$
• B. $5^6\times 15$
• C. $6^5\times \left(15\right)!$
• D. $\left(15\right)!\times 6!$

Answer 1

The correct option is D $\left(15\right)!\times 6!$

Domain $=\{1,2,3,4,...,20\}$
Co-domain $=\{1,2,3,4,...,20\}$
$\therefore k=\{4,8,12,16,20\}$
$f\left(k\right)=\{3,6,9,12,15,18\}$

Number of ways to arrange $f\left(k\right)$ for $k$
$=6\times 5\times 4\times 3\times 2=6!$

Number of ways to arrange $f\left(k\right)$ for remaining numbers $=(20-5)!=15!$
$\therefore$ Total number of functions $=6!\times 15!$