10-7-3. At a large round table, 55 people are feasting, facing the center of the table: 25 barons, 20 counts, and 10 marquises. On a signal, exactly those who have both neighbors—left and right—with the same title should stand up. What is the maximum number of people who can stand up?

For example, for a count to stand up, both of his neighbors must have the same title, but it does not necessarily have to be a count.
Answer: 52.

Solution variant 3. We will place people in a new circle every other person: $1,3,5,7$, $\ldots, 55,2,4,6, \ldots, 54$, back to 1. It is sufficient to find the maximum number of neighbors with the same title in the new circle.

It is clear that there are neighbors of different titles in the new circle. It is also clear that the number of different-title neighbors in the new circle cannot be 1. It cannot be 2 either (then the new circle would be arranged as follows: feasters of one title, then feasters of another title. Where would the third title be then?). But it can be 3: for example, first all the barons, then all the counts, then all the marquises.

Thus, we have shown that at least three out of 55 feasters will not stand up, and we have constructed an example where only three can fail to stand. Therefore, the answer to the problem is 52.