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!
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.