There are 10 photos, and the relative relationship from left to right of the three people in each photo is: son, father, father's brother. It is known that the ten people in the middle of the 10 photos are all different. How many people at least appear in these 10 photos?
Parse: Group all people by full brothers, meaning people in the same group have the same father. In 10 photos, the person on the left must be in 10 different groups because their fathers are different. The highest generation group must be different from these 10 groups, so there must be at least 511 groups. Suppose there are $n$ people in the 10 photos, then at least $11 \times 2 - n = 22 - n$ people are in groups with only one person, meaning at least $22 - n$ people have no full brothers, so they cannot be the person standing in the middle of the photos. Therefore, there are at least $22 - n + 10$ people, i.e., $22 - n + 10 \leqslant n$, so $n \geqslant 16$.
16 people can achieve this, for example, the ten photos are $3,1,2 ; 5,2,1 ; 13,7,8 ; 14, 8, 7 ; 15,11,12 ; 16,12,11 ; 10,5,6 ; 11,6,5 ; 9,4,3 ; 7,3,4.16$ people's family tree is shown in the figure below.