Question

Let $A=\{a,b,c\}$ and $B=\{1,2,3,4\}$. Then the number of elements in the set $C=\{f:A\to B|2\in f(A\left)\text{and}f\text{is not one–one}\right\}$ is

Answer 1

$C=\{f:A\to B|2\in f(A\left)\text{and}f\text{is not one–one}\right\}$
Case-I: If $f\left(x\right)=2\forall x\in A,$ then number of function $=1.$
Case-II : If $f\left(x\right)=2$ for exactly two elements then total number of many-one function.
$={}^3{C}_2\times {}^3{C}_1=9$
$\text{Case-III : If}f\left(x\right)=2\text{for exactly one elements then total number of many-one function}$
$={}^3{C}_1\times {}^3{C}_1=9$
Total $=1+9+9=19$