Permutation of different types of objects

Question

I have to arrange $8$ books, including $2$ of English, such that the books of English are never together on a shelf.

I tried to do it by first finding the total number of arrangements, which is $8!$, and subtract from it the number of arrangements if English books were kept together, which is $2!×6!×2!$.

There are $2$ English books, and they can arrange among themselves in $2!$ ways. For each of the $2!$ ways, there are $6!$ ways for the rest of $6$ books to arrange themselves and lastly, we can either keep English books first or vice versa.

The answer I got is $37440$, but the correct answer is $30240$.

I need some clarification on what I am doing wrong here.

Answer 1

Your error was in your assumption that "we can either keep English books first or vice versa." Since there are a total of eight books, the pair of English books must begin in one of the first seven positions. With that correction to your solution, we obtain $8! - 7 \cdot 6!2! = 30,240$.

Answer 2

Your method only subtracts the ways where both English books are together at the start or end of a row. However there could be any number from $0$ to $6$ books before the two English books, so you need to subtract $7\times 2!\times 6!$ instead.

Answer 3

You can view "Two English books together" as a single object/book. Then there are $7$ objects left to arrange, which is simply $7!$. English books themselves can swap, which is $2!$. In conclusion, by subtracting the unwanted ways, you have $8!-2!7!=30240$.