Problem about eight different books randomly put on shelf. The problem sounds like this (My translation from Russian): 

Eight different books were put on a shelf in random order. Calculate
  probability that two specific books were put near each other.

My answer: Let's divide the space on the shelf into eight slots. Let's also name our two books, "A" and "B" respectively.
We have two sets of combinations - in the first set of combinations we have AB (i.e. A goes first). For example, A is put into the first slot and B is put into the second slot. Next example, A is put into the second slot and B is put into the third slot. And so on. There are 7 such AB combinations in total. By the same logic there are also 7 BA combinations. Obviously there is no overlap between said combinations, thus we can sum them up and get 14 combinations in total where books A and B are put side by side.
As for number of total combinations of books on the bookshelf, it's equal to "n!", where n is equal to 8. Why? Because in order to calculate combinations when repetitions are forbidden and order is important we use this formula: 

n means total number of items and r means number of selected items. 
But because in our case n=r we get (n-r)!=0!=1. Consequently, the formula is turned into "n!".
This all means that the probability of A and B being beside each other is 14/8!
What my textbook says: My textbook has different opinion. Namely, for some strange reasons it thinks that the probability is (7*2!*6!)/8!
UPDATE:
I understand my mistake now. I forgot that while A and B can stand still in their slots we can get additional combinations by making other books to change their slots. Thus each case with positions of A and B is in fact set of combinations. How many combinations in each set? It's "6!", because we decreased number of total and selected books by ignoring books A and B. We multipy it by 14 and get 6!*14=6!*2*7=6!*2!*7
Now I'm with agreement with my textbook.
 A: Perhaps a different approach will clarify.
For the two specific books, keep one. The remaining 7 books have $7!$ possible orderings.  So to be next to the other specific book, the one kept behind has 2 positions (left and right).  This odds against all $8!$ orderings of all of the books.
$$
\frac{2 \times 7!}{8!} = \frac1{4}
$$
A: The answer is (7!.2!.)/8!
Let me clear, first consider the two book as one, (plus) the other 6 books, so we have 7 books which can put the shelf 7! different ways. Now there are two points the first one the two books now together which we want. The second point is the two books can change their places, it means 2! different ways. So we multiply them. We need divede them all possible ways. The all possible way is 8!
A: The number of options to put the books on the shelf in any order is $8!$, that is correct. Now, what is the number of options to put the books in an order such that $A$ and $B$ are next to each other? This is the way it should be done: think about $A$ and $B$ as about one book. It makes sense because you anyway need them to be next to each other. Now you need to put just $7$ "books" on the shelf-$6$ books which are not $A$ and $B$ and that "$AB$ book". So the number of ways to order them is $7!$. But then you also need to choose which of the books $A$ and $B$ will be on the left side and which will be on the right side, that is two options. So in general the number of ways to order the books where $A$ and $B$ are next to each other is $7!\times 2$, or as it appears in your textbook $7\times 2!\times 6!$. So the probability is really $\frac{7!\times 2}{8!}$. 
