Let $\mathrm{A}=\left\{{\mathrm{x}}_1,{\mathrm{x}}_2,\dots {\mathrm{x}}_7\right\} \mathrm{and} \mathrm{B}=\left\{{\mathrm{y}}_1,{\mathrm{y}}_2,{\mathrm{y}}_3\right\}$ be two sets containing seven and three distinct elements respectively. Then the total number of functions $\mathrm{f}: \mathrm{A}\to \mathrm{B}$ that are onto, if there exist exactly three elements x in A such that $\mathrm{f}\left(\mathrm{x}\right)= {\mathrm{y}}_2$, is equal to 
 

$\mathrm{A}=\left\{{\mathrm{x}}_1,{\mathrm{x}}_2,\dots {\mathrm{x}}_7\right\} \mathrm{and} \mathrm{B}=\left\{{\mathrm{y}}_1,{\mathrm{y}}_2,{\mathrm{y}}_3\right\}$

$n\left(A\right)=7,n\left(B\right)=3$

$\mathrm{Exactly} 3 \mathrm{elements} \mathrm{in} \mathrm{A} \mathrm{have} {\mathrm{y}}_2\mathrm{as} \mathrm{image} \mathrm{can} \mathrm{be} \mathrm{selected} \mathrm{in}{ }^7{\mathrm{C}}_3 \mathrm{ways}$

equal to number of onto functions from a set containing 4 elements to a set containing 2 elements=$2^4-2=14$

Number of on to functions =${ }^7{\mathrm{C}}_3\times 14$