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1.Relation and Function
hard
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
$14.{}^7{C_3}$
B
$16.{}^7{C_3}$
C
$14.{}^7{C_2}$
D
$12.{}^7{C_2}$
(JEE MAIN-2015)
Solution
Number of onto function such that exactly three elements in $x \in A$ such that $f\left( x \right) = \frac{1}{2}$ is
equal to $ = {\,^7}{C_3},\left\{ {{2^4} – 2} \right\} = 14.{\,^7}{C_3}$
Standard 12
Mathematics
