Let  $A=\{x_1,x_2,x_3\mathrm{......}{x}_7\}$,   $B=\left\{y_1{y}_2{y}_3\right\}$  The total number of functions  $f:A\to B$ that are onto and there are exactly three elements  $x$ in A such that $f\left(x\right)=y_2$, is equal to

 $A=\{x_1,x_2,x_3,x_4,x_5,x_6,x_7\},B=\{y_1,y_2,y_3\}$
$f:A\to B$ is onto  $∍f\left(x\right)=y_2$
Exactly 3 elements  $x$ in is $y_2$.  This can be done 
In ${ }^7{C}_3$  ways 
Remain A four elements in B 2 elements
 $\therefore 2^4-{}^2{C}_1{(2-1)}^4=14$
Total no.of onto functions  $={}^7{C}_3\times 14$