Solution:

The answer is (D). For convenience, let us imagine the ballerinas and the male dancers arranged horizontally, from left to right. The condition in the text can be translated as follows. Let us call $B_{i}$ the set of male dancers assigned to the $i$-th ballerina, with $i=1, \ldots, 7$. In order to respect the condition in the text (the fact that the segments do not cross), the male dancers in $B_{2}$ must all be to the right of those in $B_{1}$ (except that the rightmost male dancer of $B_{1}$ may coincide with the leftmost male dancer of $B_{2}$), and so on: the male dancers in $B_{3}$ are all to the right of those in $B_{2}$ (except that the rightmost one of $B_{2}$ may coincide with the leftmost one of $B_{3}$), those of $B_{4}$ to the right of those of $B_{3}$, and so forth. In particular, each set $B_{i}$ intersects at most one element of $B_{i-1}$ and at most one element of $B_{i+1}$ (provided $i-1 \geq 1$ and $i+1 \leq 7$). The total number of male dancers assigned, that is, the total number of segments drawn, is
$$13=|B_{1}|+|B_{2}|+\cdots+|B_{7}| \leq |B_{1} \cup B_{2} \cup \cdots \cup B_{7}|+6,$$
where the inequality holds precisely because the male dancers shared among the various sets are at most 6 (one between $B_{1}$ and $B_{2}$, one between $B_{2}$ and $B_{3}$, $\ldots$). In particular, $|B_{1} \cup \cdots \cup B_{7}|$ must be greater than or equal to 7, that is, every male dancer is assigned to at least one ballerina. The equality $13=|B_{1}|+|B_{2}|+\cdots+|B_{7}|$ also implies that $B_{1}$ actually has an element in common with $B_{2}$, which has an element in common with $B_{3}$, and so on up to $B_{7}$ (this in particular also justifies the fact, implicitly used in the previous description, that none of the sets $B_{i}$ is empty).

Let us further observe that a set $B_{i}$ cannot have 'gaps', that is, it is formed by a row of adjacent male dancers (precisely: if it contains two male dancers, it also contains all those in between them). Indeed, if two male dancers are both assigned to the same ballerina, then all the male dancers between them can only be assigned to that same ballerina (in order to respect the condition that the segments do not cross). Since we have already observed that every male dancer must be assigned to at least one ballerina, we conclude, as desired, that each set $B_{i}$ consists of mutually adjacent male dancers.

On the basis of all these observations we then obtain that to the first ballerina are assigned all the male dancers from the first to the $i_{1}$-th (for some $i_{1} \in \{1, \ldots, 7\}$); to the second, all those from the $i_{1}$-th to the $i_{2}$-th (for some $i_{2} \in \{1, \ldots, 7\}$), and so on up to the seventh ballerina, to whom are assigned the male dancers from $i_{6}$ to $i_{7}=7$. To describe the situation it is therefore enough to know the numbers $i_{1} \leq i_{2} \leq \cdots \leq i_{6}$ (since we already know that $i_{7}=7$). We can think of such a collection of numbers as a function $f$ from $\{1,2,3,4,5,6\}$ to $\{1,2,3,4,5,6,7\}$, with the condition that it be weakly increasing, that is, that $f(x+1) \geq f(x)$ for $x=1, \ldots, 6$. Knowing $f$ is then equivalent to knowing the function $g$ that sends $x$ to $f(x)+x$, which is now strictly increasing (that is, $g(x+1)>g(x)$) and takes values in the set $\{2,3, \ldots, 7+6=13\}$. Finally, given the condition of being strictly increasing, such a function is completely determined by knowledge of the 6 elements of its image (given 6 distinct numbers in $\{2, \ldots, 13\}$, there is a unique strictly increasing function $\{1, \ldots, 6\} \rightarrow \{2, \ldots, 13\}$ that assumes them as values). We conclude therefore that the number sought is the number of subsets of $\{2, \ldots, 13\}$ having 6 elements, that is $\left(\begin{array}{c}12 \\ 6\end{array}\right)=924$.