Question

Let $A=\left\{x_1,x_2,x_3,....x_7\right\},B=\{y_1,y_2,y_3.\}$ The total number of functions $f:A\to B$ such that every element in B has atleast one pre image 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 '$A$' can be selected in ${}^7{C}_3$ ways. Their image has to be $y_2.$ Remaining $2$ images can be assigned to remaining 4 pre-images in $2^4$ ways. But set B has atleast one pre image for every element of it, hence the number of ways is $2^4-2.$ Then the total number of functions is ${}^7{C}_3\times 14=490.$