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

We are asked to count all onto (surjective) functions $$f : A \rightarrow B$$ under the extra condition that exactly three elements of $$A$$ go to the element $$y_2 \in B$$. We begin by recalling what an onto function means. A function $$f$$ is onto when every element of the codomain $$B$$ has at least one pre-image in the domain $$A$$, i.e. for each $$y \in B$$ there exists at least one $$x \in A$$ with $$f(x) = y$$.

First we handle the condition about $$y_2$$. We are told that there are exactly three elements $$x$$ in $$A$$ such that $$f(x) = y_2$$. To choose which elements of $$A$$ they are, we compute

$$ ^7C_3 $$

because $$A$$ contains seven distinct elements and we must select three of them. After this choice is made, these three elements are permanently assigned the image $$y_2$$.

We now investigate what happens to the remaining elements of $$A$$. Once three elements are fixed for $$y_2$$, there are $$7 - 3 = 4$$ elements of $$A$$ left. Call this leftover set $$A'$$. These four elements must be mapped into the remaining two elements of $$B$$, namely $$y_1$$ and $$y_3$$.

Because the overall function is required to be onto, each of $$y_1$$ and $$y_3$$ must appear at least once as a value of $$f$$. So, when distributing the four elements of $$A'$$, neither $$y_1$$ nor $$y_3$$ can be missed. We count these distributions carefully.

For each of the four elements in $$A'$$ we have two immediate choices: send it to $$y_1$$ or send it to $$y_3$$. Ignoring the “at least once” restriction for the moment, this yields $$2^4$$ possibilities. Now we subtract the two unacceptable allocations—either “all go to $$y_1$$” or “all go to $$y_3$$”. Hence the number of admissible ways is

$$ 2^4 - 2 \;=\; 16 - 2 \;=\; 14. $$

Putting the two independent selections together—first the choice of which three elements map to $$y_2$$, and then the admissible distribution of the remaining four elements between $$y_1$$ and $$y_3$$—we multiply the counts:

$$ \text{Total onto functions} \;=\; {^7C_3} \times 14. $$

Thus the required number is $$14 \cdot {^7C_3}$$.

Hence, the correct answer is Option C.