Seven, the 25 knights of King Arthur sat at their round table, and three knights (any of whom were chosen with equal probability) were selected to slay the dragon. Let $\mathrm{p}$ be the probability that at least two of the three selected knights are seated next to each other. If $\mathrm{p}$ is written as a reduced fraction, find the sum of the numerator and the denominator.
The number of ways to choose 3 people from 25 is
$$
C_{25}^{3}=\frac{25 \cdot 24 \cdot 23}{3 \cdot 2 \cdot 1}=25 \cdot 23 \cdot 4 \text { (ways). }
$$

The number of ways to choose 3 people from 25 such that at least 2 are adjacent can be calculated as follows: treat the adjacent two people as one, then there are 25 ways to choose, and the other person can be chosen in 23 ways. However, this method counts the cases where all three are adjacent twice, and there are 25 ways to choose three people who are all adjacent. Therefore, the number of ways to choose 3 people from 25 such that at least 2 are adjacent is
$$
25 \cdot 23-25=25 \cdot 22 \text { (ways). }
$$

Thus, the required probability is $\frac{25 \cdot 22}{25 \cdot 23 \cdot 4}=\frac{11}{46}$.
Therefore, the sum of the numerator and the denominator is 57.