Question

Let $A=\{x_1,x_2,x_3....x_{10}\},B=\{y_1,y_2,y_3\}$. The total number of functions $f:A\to B$ that are onto and there are exactly three elements 'x' in A such that $f\left(x\right)=y_1$, is equal to

• A. ${126.}^7{C}_2$
  
• B. ${126.}^{10}{C}_3$
  
• C. ${10.}^7{C}_2$
  
• D. ${7.}^7{C}_3$

Answer 1

The correct option is B ${126.}^{10}{C}_3$


$A=\{x_1,x_2,x_3,x_4,x_5,x_6,x_7\},B=\{y_1,y_2,y_3\}$
$f:A\to B$ is onto such that $f\left(x\right)=y_1$
Exactly 3 elements x in is $y_1$. This can be done in ${10}_{C_3}$ ways
Remaining A has seven elements and B has 2 elements
$\therefore 2^7{-}^2{C}_1(2-1{)}^7=126$
Total no.of onto functions ${=}^{10}{C}_3\times 126$