Final answer:
To organize 8 different books (3 math books, 3 history books, and 2 novels) without any two math books next to each other, there are 4,320 possible arrangements.
Step-by-step explanation:
To organize 8 different books (3 math books, 3 history books, and 2 novels) without any two math books next to each other, we can use the concept of permutations. We have a total of 8 books, so there are 8! (eight factorial) ways to arrange them initially.
However, since no two math books can be next to each other, we need to consider the restriction. Let's assume the math books are labeled M1, M2, and M3. We can treat these three math books as a single entity, and there are 3! ways to arrange them within this entity. Within this entity, the math books could be arranged as M1M2M3, M1M3M2, M2M1M3, M2M3M1, M3M1M2, or M3M2M1.
Now, we have 6 entities (3 math books, 3 history books, and 2 novels) to arrange. There are 6! ways to arrange these entities.
Finally, we multiply the arrangements within the math book entity (3!) by the arrangements of the remaining entities (6!), giving us a total of 3! x 6! = 6 x 720 = 4,320 possible arrangements.