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 $f(x)=y_2$, is equal to :

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

Answer 1

Answer: (D)

$A\left\{x_1,x_2,x_3,\dots x_7\right\};B=\left\{y_1,y_2,y_3\right\}3$ elements in $A$ having image $y_2$ can be chosen in ${}^7{C}_3$ ways. Now we are left with 4 elements in $A$ which are to be associated with $y_1$ or $y_3$ i.e.
each of 4 elements $A$ has 2 choices $y_1$ or $y_3$ i.e. in $(2{)}^4$ ways.
But there are 2 ways when one element of $B$ will remain associated i.e.
when all 4 are associated with $y_1$ or $y_3$.
$\therefore$ Regd. No. of functions $={}^7{C}_3\left((2{)}^4-2\right)$
$=14\cdot \left({}^7{C}_3\right)$