The number of functions f from set $A=\{1,2,3,\dots ,20\}$ onto itself satisfying f(k) = multiple of 3 ,whenever it is a multiple of 4. is

An onto function from A to itself is a bijection A multiple of 4 in domain A can he 4, 8, 12, 16, 20 and a multiple of 3 in co-domain can take 6 values 3, 6, 9, 12, 15, 18.

 It is given that the image of a multiple of 4 is a multiple of 3 and no two multiples of 4 can have the same image. So, the number of ways of associating multiples of 4 to multiples of $3$is ${ }^6{C}_5\times 5!$ 

Remaining 15 elements in domain can be associated to 15 elements in co-domain in 15! ways. 

Hence, the required number of functions is : ${ }^6{C}_5\times 5!\times 5!=6!\times 15!$