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1.Relation and Function
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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
$14{(^7}{C_3})$
Solution
Number of ways of selection of three elements in $A$ such that
$f(x) = {y_2}$ is $^7{C_3}$
Now, for remaining $4$ elements in $A,$ we have $2$ elements is $\mathrm{B}$
$\therefore $ Total number of onto functions
$ = {\,^7}{{\rm{C}}_3} \times \left( {{2^4} – {\,^2}{{\rm{C}}_1}(2 – 1)} \right) = 14\left( {^7{{\rm{C}}_3}} \right)$
Standard 12
Mathematics
