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In how many ways can eight books, including two of English, can be arranged in such a way that the English books are never together (i. e. next to each other)?

My Attempt:

There are a total of $8! = 40320$ ways of arranging those eight books.

First, we consider the case when the two English books are next to each other. Then these two books may be regarded as constituting a single block, and this block can appear in seven different positions, thus making for seven mutually exclusive cases.

In each of these seven cases, there are two ways of arranging the two books on English within the block.

What next? How to proceed from here?

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    $\begingroup$ In the second case, there are $2\cdot7!$ possibilities, for the reasons you have given. $\endgroup$
    – saulspatz
    Jul 13 '21 at 13:46
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As saulspatz notes, since there are $7!$ ways of ordering the books with the two English books together as an indistinguishable block, and therefore $2\cdot 7!$ ways of ordering the books where the English books are together, looking at the books separately, and since there are $8!$ orderings in total, there are $8!-2\cdot 7!$ orderings with the books NOT together, and thus apart, as the books either are or aren’t together so we must be able to partition the orderings this way.

$$8!-2\cdot 7!=8\cdot 7!-2\cdot 7!=(8-2)\cdot 7!=6\cdot 7!=30,240$$

If my mental maths holds true.

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