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 $f : A \rightarrow B$ 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)

$A\left\{x_{1}, x_{2}, x_{3}, \ldots x_{7}\right\} ; B=\left\{y_{1}, y_{2}, y_{3}\right\} 3$ 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 $={ }^{7} C_{3}\left((2)^{4}-2\right)$
$=14 \cdot\left({ }^{7} C_{3}\right)$