Let $A = \left\{ {{x_1},{x_2},{x_3},.....,{x_7}} \right\}$ and $B = \left\{ {{y_1},{y_2},{y_3}} \right\}$ be two sets containing seven and three distinct elements respectively. Then the total number of functions $f:A \to B$ which 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

    $14{(^7}{C_3})$

Similar Questions

Let $\quad E_1=\left\{x \in R : x \neq 1\right.$ and $\left.\frac{x}{x-1}>0\right\}$ and $\quad E_2=\left\{x \in E_1: \sin ^{-1}\left(\log _e\left(\frac{x}{x-1}\right)\right)\right.$ is a real number $\}$.

(Here, the inverse trigonometric function $\sin ^{-1} x$ assumes values in $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ )

Let $f : E _1 \rightarrow R$ be the function defined by $f(x)=\log _c\left(\frac{x}{x-1}\right)$ and $g: E_2 \rightarrow R$ be the function defined by $g(x)=\sin ^{-1}\left(\log _e\left(\frac{x}{x-1}\right)\right)$

 $LIST I$ $LIST II$
$P$ The range of $f$ is $1$ $\left(-\infty, \frac{1}{1- e }\right] \cup\left[\frac{ e }{ e -1}, \infty\right)$
$Q$ The range of $g$ contains $2$ $(0,1)$
$R$ The domain of $f$ contains $3$ $\left[-\frac{1}{2}, \frac{1}{2}\right]$
$S$ The domain of $g$ is $4$ $(-\infty, 0) \cup(0, \infty)$
  $5$ $\left(-\infty, \frac{ e }{ e -1}\right]$
  $6$ $(-\infty, 0) \cup\left(\frac{1}{2}, \frac{ e }{ e -1}\right]$

The correct option is:

  • [IIT 2018]

Let $f(x)=a x^{2}+b x+c$ be such that $f(1)=3, f(-2)$ $=\lambda$ and $f (3)=4$. If $f (0)+ f (1)+ f (-2)+ f (3)=14$, then $\lambda$ is equal to$...$

  • [JEE MAIN 2022]

If $\,\,f(x) = \left\{ {\begin{array}{*{20}{c}}
  {3 + x;\,\,\,\,\,x \geqslant 0} \\ 
  {2 - 3x;\,\,\,\,\,x < 0} 
\end{array}} \right.$ then $\mathop {\lim }\limits_{x \to 0} f(f(x))$ is equal to -

If $f$ is an even function defined on the interval $(-5, 5)$, then four real values of $x$ satisfying the equation $f(x) = f\left( {\frac{{x + 1}}{{x + 2}}} \right)$ are

Range of ${\sin ^{ - 1\,}}\left( {\frac{{1 + {x^2}}}{{2 + {x^2}}}} \right)$ is