The number of onto functions f from {1, 2, 3, | vedclass

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(C) Let $S = \{1, 2, 3, \dots, 20\}$. The elements in $S$ that are multiples of $4$ are $K = \{4, 8, 12, 16, 20\}$. There are $5$ such elements.For $k

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$15! 	imes 6!$

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(C) Let $S = \{1, 2, 3, \dots, 20\}$. The elements in $S$ that are multiples of $4$ are $K = \{4, 8, 12, 16, 20\}$. There are $5$ such elements.For $k \in K$,$f(k)$ must be a multiple of $3$. The multiples of $3$ in $S$ are $M = \{3, 6, 9, 12, 15, 18\}$. There are $6$ such elements.Since $f$ is an onto function,the $5$ elements of $K$ must be mapped to $5$ distinct elements of $M$. The number of ways to choose and arrange these is $^6P_5 = \frac{6!}{1!} = 6!$.The remaining $15$ elements of $S \setminus K$ must be mapped onto the remaining $15$ elements of $S \setminus f(K)$ in a bijective manner,which can be done in $15!$ ways.Therefore,the total number of onto functions is $6! 	imes 15!$.

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

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