Question

Let $A=\{x_1,x_2,.......,x_7\}$ and $B=\{y_1,y_2,y_3\}$ be two sets containing seven and three distinct elements respectively. Then the total number of functions $f:A\to B$ that are onto. If there exist exactly three elements $x$ in $A$ such that $f\left(x\right)=y_2$, is equal to:

• A. ${12.}^7{C}_2$
• B. ${16.}^7{C}_3$
• C. ${14.}^7{C}_3$
• D. ${14.}^7{C}_2$

Answer 1

The correct option is C ${14.}^7{C}_3$

Given, there are exactly three elements in $A$ such that $f\left(x\right)=y_2$..

Number of ways of selecting $3$ elements from $A={}^7{C}_3$

Now the each remaining element in $A$ can be associated with either of $y_1,y_3$ in $B$.

Total No of ways = $2\times 2\times 2\times 2$, but subtracting the cases where all the remaining elements getting associated to either $y_1$ or $y_3$ , we have $16-2=14$ ways.

Therefore total number of onto functions = $14\cdot {}^7{C}_3$