In a company, there are $n$ married couples together, and it turns out that the people present form $n$ sibling pairs. In how many ways is this possible for the cases where $n \leq 5$? (Siblings, of course, cannot be married to each other.)
I. Solution. For every person present, there are 2 other people with whom they have a relationship relevant to the question, one being a marital partner (indicated by a double line in the diagram) and the other a sibling (indicated by a single line). Let $t_{n}$ denote the number of possible (e.g., guessable) cases for $n$ married couples.

Let's start our reasoning with the oldest member of the group, $A$ (who can be uniquely identified), let $a$ be their spouse, and $b$ their sibling. Furthermore, let $B$ be $b$'s spouse (1st figure). These are clearly all different individuals, so $n \geq 2$, hence $t_{1}=0$.

For $n=2$, there are no more people present, so the second sibling couple can only be $B, a$, but since $b$ could be either member of the second couple, $t_{2}=2$.

If $n=3$, then $B$ and $a$ cannot be siblings, otherwise the third couple would also be a sibling couple. Thus, $B$'s sibling is one of the members of the third couple, denoted by $c$, and their spouse by $C$, and in this case, $C$'s sibling can only be $a$. In our current line of thought, we can guess $b$ in 4 different ways from the 2 couples remaining after $A, a$, and we can guess $c$ in 2 different ways, so $t_{3}=4 \cdot 2=8$ (2nd figure).

$$
\underbrace{a=A-b=B}_{\text {1st figure }} \quad \underbrace{a=A-b=B-c=C}_{\text {2nd figure }} \quad \underbrace{c=C-d=D}_{\text {3rd figure }}
$$

For $n=4$, $B$'s sibling can again be $a$, or a different member of the group from the ones already mentioned $(A, a, b, B)$, let this be $c$.

In the former case, the first two couples provide a solution similar to the case of $n=2$, and the same applies to the two remaining couples, the possibilities for the latter can be counted starting from the oldest among them, $C$, the group can then be arranged in 2 circles (1st and 3rd figures).

In the latter case, $C$'s sibling is again not $a$, but one of the members of the 4th couple, let this be $d$, then the spouse of this, $D$, forms a sibling pair with $a$; in this case, the diagram describing the guess is longer by the ",-d=D" part compared to the 2nd figure, and again, all present can be arranged in 1 circle. Now, there are always 6 possible guesses for $b$, and further, for $d$, $t_{2}$, i.e., 2 guesses (since in this case we do not choose $c$, but it is determined), and in the latter case, 4 for $c$ and 2 for $d$, so the continuation is $4 \cdot 2$, thus $t_{4}=6(2+4 \cdot 2)=60$.

Finally, for $n=5$, following the pattern of the previous cases, the sibling of either $B$, $C$, or a member of the 5th couple $E$. In the first and second cases, a new circle is formed from the remaining 3, or 2 couples, according to the case of $n=3$, or $n=2$; there are always 8 possibilities for $b$, $c$ either is determined or can be chosen from 6 people, and in the latter case, $d$ is either determined or can be chosen from 4, and $e$ from 2:

$$
t_{5}=\left[t_{3}+6\left(t_{2}+4 \cdot 2\right)\right]=8(8+60)=544
$$

II. Solution. Let's denote the married couples by the numbers $1,2, \ldots, n$, and the two members of the $k$-th couple by $k_{1}$ and $k_{2}$ $(1 \leq k \leq n)$; furthermore, let the number of ways to arrange the $n$ married couples into sibling pairs be $t_{n}$.

Let the sibling of $1_{1}$ be $k_{i}$, where $k \neq 1$ and $i=1$ or 2, so there are $2(n-1)$ possibilities for its selection. Consider the sibling of the spouse of $k_{i}$, $k_{3-i}$. If this is exactly $1_{2}$, then we have settled the 1st and $k$-th couples, and the number of ways to arrange the remaining $(n-2)$ couples into sibling pairs is $t_{n-2}$.

If, however, $k_{3-i}$ and $1_{2}$ are not siblings, then they can be considered a pseudo-couple, because they have the same property that we need to consider for the real couples, namely that they cannot form a sibling pair. Thus, we need to form $(n-1)$ sibling pairs from the remaining $(n-1)$ couples (real and pseudo), for which there are $t_{n-1}$ possibilities. There is no other possibility for $k_{3-i}$, so

$$
t_{n}=(2 n-2)\left(t_{n-2}+t_{n-1}\right)
$$

This reduces the determination of $t_{n}$ to the determination of $t_{n-1}$ and $t_{n-2}$, and essentially solves the problem for any natural number $n$ with the obtained so-called recursive (backtracking, relying on previous values) formula.

Now, $t_{1}=0$ and $t_{2}=2$ (since the sibling of $1_{1}$ can be either $2_{1}$ or $2_{2}$, and this determines the other sibling pair as well), based on which $t_{3}=4\left(t_{1}+t_{2}\right)=8 ; t_{4}=6\left(t_{2}+t_{3}\right)=60$; finally, $t_{5}=8\left(t_{3}+t_{4}\right)=544$.

Remark. Following the line of thought of the I. solution, we obtain the following relationship for any $n$:

$$
\begin{gathered}
t_{n}=(2 n-2) t_{n-2}+(2 n-2)(2 n-4) t_{n-3}+\ldots+(2 n-2)(2 n-4) \cdot \ldots \cdot 4 t_{1}+ \\
+(2 n-2)(2 n-4) \cdot \ldots \cdot 4 \cdot 2
\end{gathered}
$$

The value of $t_{n}$ can also be determined step by step based on this.