Question

Let $A=\{x_1,x_2,x_3,....,x_7\},B=\{y_1,y_2,y_3\}$. The total number of functions $f:A\to B$ that are on to and there are exactly three element $x$ in $A$ such that $f\left(x\right)=y_2$ is equal to

• A. $490$
• B. $510$
• C. $630$
• D. None of these

Answer 1

The correct option is A $490$

Three elements from set ${}^{\prime }{A}^{\prime }$ can be selected in ${}^7{C}_3$ ways. Their image has be $y_2$. Remaining $2$ images can be assigned to remaining $4$ pre-images in $2^4$ ways. But the function is onto, hence the number of ways is $2^4-2$. Then the total number of functions is ${}^7{C}_3\times 14=490$.