$5 \cdot 41$ In a family photo album, there are 10 photos, and each photo has 3 people: the man standing in the middle, his son standing on his left, and his brother standing on his right. If it is known that the 10 men standing in the middle in the 10 photos are all different, how many different people are there at least in these photos?
[Solution] We refer to the person standing in the middle of each photo as the main character, and the rest as supporting characters. We categorize all the people in the photos by generation: the highest generation is called the 0th generation, and when $k$ $=0,1,2, \cdots$, the sons of the $k$th generation in the photo belong to the $k+1$th generation. Let $r_{k}$ represent the number of $k$th generation main characters, and $t_{k}$ represent the number of $k$th generation supporting characters. Since each main character has a brother, and the two have a common father, the number of fathers of the $k+1$th generation men does not exceed $\frac{1}{2} r_{k+1}+t_{k+1}$. At the same time, because each main character has a son, the number of relatives of the $k+1$th generation men is no less than $r_{k}$. Therefore, we have
$$
r_{k} \leqslant \frac{1}{2} r_{k+1}+t_{k}, \quad k=0,1,2, \cdots .
$$

Additionally, $1 \leqslant \frac{1}{2} r_{0}+t_{0}$. Adding these inequalities, we get
$$
\begin{array}{c}
\frac{1}{2}\left(r_{0}+r_{1}+\cdots\right)+1 \leqslant t_{0}+t_{1}+\cdots . \\
\left(r_{0}+r_{1}+\cdots\right)+\left(t_{0}+t_{1}+\cdots\right) \geqslant \frac{3}{2}\left(r_{0}+r_{1}+\cdots\right)+1 \\
=\frac{3}{2} \times 10+1=16 .
\end{array}
$$

Thus, there are at least 16 different people in the photos.
On the other hand, when the kinship of 16 people is as shown in the diagram on the right, where horizontal lines indicate brother relationships, and diagonal and vertical lines indicate father-son relationships, the 10 photos can be arranged as follows: $(3,1,2),(5,2,1)$, $(7,3,4),(9,4,3),(11,5,6)$, $(12,6,5),(13,7,8),(14,8,7),(15,9,10),(16,10,9)$. It is easy to see that these 10 photos meet the requirements of the problem.
In summary, there are at least 16 different people in these 10 photos.