Let A = {x_1, x_2, dots, x_7} and B = {y_1, y | vedclass

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(A) We are given $A = \{x_1, \dots, x_7\}$ and $B = \{y_1, y_2, y_3\}$.We need to find the number of onto functions $f: A 	o B$ such that exactly thr

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$14 	imes {}^7C_3$

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(A) We are given $A = \{x_1, \dots, x_7\}$ and $B = \{y_1, y_2, y_3\}$.We need to find the number of onto functions $f: A 	o B$ such that exactly three elements in $A$ map to $y_2$.First,we choose $3$ elements from $A$ to map to $y_2$,which can be done in ${}^7C_3$ ways.Now,the remaining $4$ elements of $A$ must map to the remaining $2$ elements of $B$,which are $\{y_1, y_3\}$.For the function to be onto,the set $\{y_1, y_3\}$ must be covered by the $4$ elements.The total number of functions from these $4$ elements to $\{y_1, y_3\}$ is $2^4 = 16$.We must exclude the cases where all $4$ elements map to only $y_1$ or only $y_3$ to ensure the function is onto (i.e.,$y_1$ and $y_3$ are both covered).Thus,the number of ways is $2^4 - 2 = 16 - 2 = 14$.Therefore,the total number of such onto functions is ${}^7C_3 	imes 14 = 14 	imes {}^7C_3$.

Let $A = \{x_1, x_2, \dots, x_7\}$ and $B = \{y_1, y_2, y_3\}$ be two sets containing seven and three distinct elements respectively. The total number of onto functions $f : A \to B$ such that there exist exactly three elements $x$ in $A$ with $f(x) = y_2$ is equal to:

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