Thomas's Bookshelf: How Many Ways Two English Books Can Be Separated

Let's approach this problem step by step. Since there are 2 Chinese books (C), 4 English books (E), and 3 Mathematics books (M), it means we have a total of 9 books. Now, let's first calculate the number of ways to arrange these books without restriction. This will give us insight into how many ways we can arrange the books disregarding the condition that any two English books cannot be together. There are 9! (9 factorial) ways to arrange these books but since there are 2 Chinese books, 4 English books, and 3 Mathematics books that are identical, we also need to consider their permutations: 9! / (2! * 4! * 3!) Now, let's calculate the number of ways the books can be arranged if we place two or more English books together. We can treat the two adjacent English books as one item for now and replace them later. In this case, we have: - 2 Chinese books (C) - 1 pair of English books (EE) - 2 individual English books (E) - 3 Mathematics books (M) This leaves us with a total of 8 positions and 5 unique items. The way to arrange these items is: 8! / (2! * 1! * 2! * 3!) Next, we need to break the adjacent pair of English books back into individual English books. We form the pair of English books using 4 books, so there are 4! ways to arrange the two books within the EE pair. This results in the total number of ways to arrange the books with two or more English books together: (8! / (2! * 1! * 2! * 3!)) * 4! Finally, we subtract the arrangements with English books together from the total unrestricted arrangements to get the number of ways where no two English books are together: (9! / (2! * 4! * 3!)) - (4! * (8! / (2! * 1! * 2! * 3!))) Once we calculate the difference, we get the answer: (9! / (2! * 4! * 3!)) - (4! * (8! / (2! * 1! * 2! * 3!))) = 1260 So, there are 1260 ways to arrange the books such that no two of the English books are together.