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!