The number of ways of arranging 10 different books on a shelf such that two particular books are always together is:

Out of the ten books, we need to have a case where a specific pair of two books is always there. 

To solve this, the best way is to assume the two books as a single element. This leaves us with nine different elements, where one of them has two books in it. 

To arrange nine different elements, we have 9! ways.

Now, it is important to note that the two books considered as a single element can be arranged within themselves as XY or YX (2 ways). 

Therefore, we shall multiply the initial result by 2, which makes the total number of ways as 9!$$\times\ $$2.