Here, we have to find the number of ways in which we can arrange 10 books on a shelf such that a particular pair of books is never together. First let's find the number of ways in which 10 books can be arranged on a shelf. No. of ways to arrange all the 10 books on a shelf = ‌10P10=10! Now let's find out the number of ways to arrange these 10 books on a shelf such that a particular pair of books is always together. Since the particular pair of books is always together consider the pair as one book. Then we need to arrange 9 books on the shelf. This can be done in ‌9P9=9! ways. Now the pair of books which are considered as one book can be arranged in 2! ways So, number of ways to arrange these 10 books on a shelf such that a particular pair of books is always together = 2 × 9! We know that, the number of ways of arranging 10 books on a shelf so that a particular pair is never together = 10! - (2 × 9!) = 8 × 9! Hence, the required number of ways to arrange the books on shelf as per the requirement = 8 × 9!