I had no problems with a and b but c makes absolutely no sense in the back of the book, I think it has to be wrong. It is frustrating how wrong I think the book is. The question asks if the novels must be together but the other books can be arranged in any order, how many arrangements are possible?
NNN MM C
Clearly there are:
$3!$ ways that the novels can be arranged
$2!$ ways that the math text books can be arranged
$3!$ ways that MM and C can be arranged
$3!$ ways that the novels grouped together as one, the math text books, and the chemistry textbook can all be arranged.
This of course comes to 432 ways that they can be arranged but the back of the book says there are 144 ways which is substantially less. I think the book is saying that MM an C have two ways that they can be arranged but that ONLY MAKES SENSE IF math and chemistry have to remain SEPARATE, however since they do not have to, there are 3 books in this instance which of course can be arranged 3! ways. The way the book would have it is $3! \times 3! \times 2! \times 2!$
Best Answer
Consider the four blocks $[N_1N_2N_3], [M_1], [M_2]$ and $[C]$. There are $4!$ ways to arrange them. Moreover, for each such arrangement, there are $3!$ ways to arrange $N_1N_2N_3$. Hence, the total is
$$4!\times 3!=144$$