410 translators are invited to an international mathematics conference. Each translator is proficient in exactly two of the five languages: Greek, Slovenian, Vietnamese, Spanish, and German, and no two translators are proficient in the same pair of languages. The translators are to be assigned to five rooms, with two translators in each room, and these two translators must be proficient in the same language. How many different allocation schemes are there (considering all possible allocation schemes of 5 pairs of translators arranged in five rooms as the same scheme)?
Construct a graph $G$, where points $x_{1}, x_{2}, x_{3}, x_{4}, x_{5}$ represent five languages, and any two points (such as $x_{1}, x_{2}$) are connected by an edge $(x_{1} x_{2})$ representing a translator proficient in these two languages.
By the problem statement, graph $G$ is a simple complete graph.
Next, orient all edges of graph $G$ according to the following rule: if the common language of the room where the translator $x_{1} x_{2}$ is located is $x_{1}$, then the edge $x_{1} x_{2}$ is oriented as $x_{2} \rightarrow x_{1}$, which can also be denoted as $\overrightarrow{x_{2} x_{1}}$.

By the problem statement, graph $G$ is a simple directed graph, and the in-degrees of all points in graph $G$ are even (0, 2, or 4).
(1) If there are two points $x_{1}, x_{2}$ with in-degree 4, then the edge $x_{1} x_{2}$ is both $x_{1} \rightarrow x_{2}$ and $x_{2} \rightarrow x_{1}$, which is a contradiction.
(2) If the in-degree of each point is 2, then the out-degree is also 2.
Take any point $x_{1}$, and draw two edges $\overrightarrow{x_{1} x_{2}}$ and $\overrightarrow{x_{4} x_{5}}$.

As shown in Figure 2, place $x_{i} (i=1,2, \cdots, 5)$ in a circle.
$x_{3} \rightarrow x_{4}, x_{3} \rightarrow x_{5};$
$x_{2} \rightarrow x_{4}, x_{3} \rightarrow x_{4};$
From $\overrightarrow{x_{3} x_{5}}$ and $\overrightarrow{x_{4} x_{5}}$, we have
$x_{5} \rightarrow x_{1}, x_{5} \rightarrow x_{2}$.
Thus, we obtain graph $G^{\prime}$, which corresponds one-to-one with a 5-cycle arrangement.
Therefore, there are $4! = 24$ allocation schemes.
(3) If there is a point $x_{1}$ with in-degree 4, then there must be another point $x_{2}$ with out-degree 4, and the remaining three points have in-degree 2.
Clearly, there are edges
$\overrightarrow{x_{i} x_{1}} (i=2,3,4,5), \overrightarrow{x_{2} x_{j}} (j=3,4,5)$.
Thus, $x_{3}, x_{4}, x_{5}$ form a unidirectional cycle, which corresponds one-to-one with a 3-cycle arrangement, and $\overrightarrow{x_{i} x_{1}}$ can be divided into two groups in three ways.
Therefore, there are $5 \times 4 \times 2! \times 3 = 120$ allocation schemes.
In summary, there are 144 allocation schemes.