Domain and codomain ={1,2,3,.....,20}. 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, when ever k is multiple of 4 then f(k) is multiple of 3 then total number of arrangement =6c5×5!=6! Remaining 15 elements can be arranged in 15! ways. Since, for every input, there is an output ⇒ function f(k) in onto ∴ Total number of arrangement =15!.6!