6.5 Arrange 5 different red beads and 3 different blue beads around a circular plate. How many ways are there to arrange them? If the blue beads are not adjacent, how many ways are there? What if the blue beads are together?
6. This is a circular permutation, 8 beads arranged in a circle is $\frac{8!}{8}=7!$.

If the blue beads are not adjacent; first consider the 5 red beads which have 4! permutations, then insert the blue beads one by one, the first one has 5 possibilities, the second one has 4 possibilities, the third one has 3 possibilities, thus there are $4!\times 5 \times 4 \times 3=1440$.

If the blue beads are all adjacent, consider the 3 blue beads as one, together with the 5 different red beads making a total of 6, the circular permutation is 5!. Since the 3 blue beads are different, there are 5!3! in total.
7. Let the regular $(2n+1)$-gon be $A_{1} A_{2} \cdots A_{2 n+1}$, first consider $\triangle A_{1} A_{i} A_{j}$ with $A_{1}$ as one vertex, the second vertex $A_{i} \in\left\{A_{2}, A_{3}, \cdots, A_{n+1}\right\}$, after $A_{i}$ is determined, the third point $A_{j} \in$ $\left\{A_{n+2}, \cdots, A_{n+i}\right\}$, hence there are $\sum_{i=2}^{n+1}(i-1)=\frac{n(n+1)}{2}$ kinds, the total number of acute triangles is $\frac{2 n+1}{3} \cdot \frac{n(n+1)}{2}$.