Let A = {x_1, x_2, x_3, dots, x_7} and B = {y | vedclass

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(D) We are given that $f: A 	o B$ is an onto function where $|A| = 7$ and $|B| = 3$. It is given that exactly three elements in $A$ map to $y_2 \in B

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$14(^7C_3)$

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(D) We are given that $f: A 	o B$ is an onto function where $|A| = 7$ and $|B| = 3$. It is given that exactly three elements in $A$ map to $y_2 \in B$. The number of ways to choose these $3$ elements from $7$ is $^7C_3$. Now,the remaining $4$ elements in $A$ must map to the remaining $2$ elements in $B$ (i.e.,${y_1, y_3}$) such that the function is onto. For the function to be onto,both $y_1$ and $y_3$ must be mapped to by at least one element from the remaining $4$ elements. The total number of functions from a set of $4$ elements to a set of $2$ elements is $2^4 = 16$. The number of functions that are not onto (i.e.,mapping to only one element) is $^2C_1 = 2$. Thus,the number of onto functions from the remaining $4$ elements to ${y_1, y_3}$ is $2^4 - 2 = 16 - 2 = 14$. Therefore,the total number of such onto functions is $^7C_3 	imes 14 = 14(^7C_3)$.

Let $A = \{x_1, x_2, x_3, \dots, x_7\}$ and $B = \{y_1, y_2, y_3\}$ be two sets containing seven and three distinct elements respectively. Then the total number of functions $f: A \to B$ which are onto,if there exist exactly three elements $x$ in $A$ such that $f(x) = y_2$,is equal to

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