Let $\mathrm{A}=\left\{{\mathrm{x}}_1,{\mathrm{x}}_2,{\mathrm{x}}_3,\dots \dots {\mathrm{x}}_8\right\},\mathrm{B}=\left\{{\mathrm{y}}_1,{\mathrm{y}}_2,{\mathrm{y}}_3\right\}$ the total number of functions $\mathrm{f} : \mathrm{A}\to \mathrm{B}$ that are onto and there are exactly four elements (x) in A such that $\mathrm{f}\left(\mathrm{x}\right)={\mathrm{y}}_3$, is equal to

Four elements from set A can be selected in ${ }^8{\mathrm{C}}_4$ ways their image will be ${\mathrm{y}}_3$. Now number of ways in which four elements of A can be connected with two elements of B in $2^4-2=14\text{ ways }$

Required number of functions${=}^8{\mathrm{C}}_4\cdot 14$