Total number of ways 10 books can be arranged if 4 specific books may not be together? I am trying to solve this albeit probably juvenile combination/permutation type question.
A person has 10 books, of which four are fiction books. How many ways can the 10 books be arranged so that all four fiction books are not together?
My reasoning thus far is this:
(Total number of ways without the restriction) minus (Total with books together)
i.e. 10! - (6! x 4!)

Am I on the right track? Thanks!
 A: Let's suppose an "arrangement" is a determination of a linear order of the $n=10$ unique books, so that there are $n!$ possible arrangements.  In any such arrangement the $n-k=10-4=6$ non-fiction books will appear in some order and there are $$(n-k)!$$ ways to do that.  Independently of that arrangement of the non-fiction books, the fiction books have to be placed either before, between, or after the non-fiction books: there are $n-k+1 = 7$ such places.  No such place may be occupied by more than one fiction book.  There are $$\binom{n-k+1}{k}$$ choices of $k$ of those places in which to put the fiction books and $$k!$$ distinct arrangements of the fiction books for each such choice.
The answer therefore is
$$(n-k)!\binom{n-k+1}{k}k! = \frac{(n-k)!(n-k+1)!}{(n-2k+1)!}.\tag{1}$$

Another way to solve this is to note that when the books are arranged vertically on top of one extra non-fiction book, then every fiction book lies on top of a non-fiction book.  That will appear as $k=4$ pairs of fiction-nonfiction books interspersed between $n-2k+1=3$  nonfiction books, for which there are therefore $$\binom{n-2k+1+k}{k}=\binom{n-k+1}{k}$$ possibilities.  That implies the chance of seeing such a pattern within the set of all $n!$ possible arrangements is the ratio $$\frac{\binom{n-k+1}{k}}{\binom{n}{k}}.$$  Multiply this by $n!$ to get the answer $$\frac{\binom{n-k+1}{k}}{\binom{n}{k}}n!\tag{2}.$$

As an example, let $n=5$ books of which $k=2$ fiction books must be kept separated.  The formulae $(1)$ and $(2)$ give
$$\frac{(5-2)!(5-2+1)!}{(5 - 2(2) + 1)!} = \frac{3!4!}{2!}=72\tag{1a}$$ 
and 
$$\frac{\binom{5-2+1}{2}}{\binom{5}{2}}5! = \frac{6}{10}120=72\tag{2a}.$$
In light of the second analysis, it suffices to illustrate this by showing all the valid patterns of the five books, writing "N" for non-fiction and "F" for fiction:
$$NNNFF, \color{red}{NNFNF}, \color{red}{NFNNF}, \color{red}{FNNNF}, NNFFN, \\\color{red}{NFNFN}, \color{red}{FNNFN}, NFFNN, \color{red}{FNFNN}, FFNNN$$
The red ones are the $\binom{4}{2}=6$ valid patterns out of all $\binom{5}{2}=10$ possible patterns.
As another example, let there be $k=3$ fiction books out of these five.  Now only one of the ten patterns is valid -- $FNFNF$ -- and indeed $1/10=\binom{3}{3}/\binom{5}{3}$.  Consequently there are only $120/10=12$ such arrangements.  They are found by applying any of the $3!=6$ permutations of the fiction books and $2!=2$ permutations of the non-fiction books within this arrangement.
A: You are on the right track, only that instead of 6! you should use 7!, because you have 7 entities - the other 6 books (non fiction) and the group of 4 fiction books that, although they should be together (in the complementary event), can still be moved as a group anywhere among the other 6 books. So in my opinion, if I understand the question correctly, the answer is
10! - (7! x 4!)
