Question

The number of functions f from {1, 2, 3…….19, 20} onto {1, 2, 3…….19, 20} such that f(k) is a multiple of 3, whenever k is a multiple of 4, is
a)$15! \times 6!$
b) ${5}^{6} \times 15$
c) $5! \times 6!$
d) ${6}^{5}\times 15!$

Answer 1

We know that a function is said to be onto function if every element of that function has its image on the other function. If the number of elements in a set is n then the arrangement of n elements can be done in $n!$ ways.

Complete step-by-step solution:
Let set $A = {1, 2, 3…….19, 20}$
As given in the question $f(k)$ is a multiple of 3, so,
$f(k)= {3, 6, 9, 12, 15, 18}$
the number of elements in set $f(k)=6$
as given in the question k is multiple of 4, so,
$k= {4, 8, 12, 16, 20}$
then the number of elements in set $k =5$
So, k to be an onto function every element of k should have its unique image inset f(k).
We know that number of elements in k is and number of element in f(k) is 6 , so, selection of images of 5 elements of set k from 6 element of set f(k) is given by
$\Rightarrow {}^{n}{{C}_{r}}$
Where $n=6$ and $r=5$
$\Rightarrow {}^{6}{{C}_{5}}$
The images can be arrange in 5! Ways, so,
$\Rightarrow {}^{6}{{C}_{5}}\times 5!$
We know that ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$
$\begin{aligned} & \Rightarrow {}^{6}{{C}_{5}}\times 5!=\dfrac{6!}{5!\left( 6-5 \right)!}\times 5! \\\ & \Rightarrow 6! \\\ \end{aligned}$
As we have calculated only for 5 elements so 15 elements are remaining. The number of onto functions for remaining 15 =$15!$
So, the total number of functions is 15! \times 6!
So, the answer is a)$15! \times 6!$

Note: We know that a function is said to be onto function if every element of that function has its image on the other function. If each element of set B does not have its pre-define image in set A then it is not onto function. The onto functions are also known as Surjective Functions