number of ways to arrange books

Question

I am solving number of arrangements for following question:

Eight books are placed on a shelf. Three of them form a 3-volume series, two form a 2-volume series, and 3 stand on their own. In how many ways can the eight books be arranged so that the books in the 3-volume series are placed together according to their correct order, and so are the books in the 2-volume series? Noted that there is only one correct order for each series.

I analyzed this problem in following way: 3 volume book as A, 2 volume book as B, and remaining books as C. So three can be arranged as 3!, and C in turn can be arranged as 3! which by product rule we can have 3! * 3! = 36. Why this analysis or approach to this problem is wrong. Correct answer is 120 (i.e. 5!). Kindly help. thanks

Answer 1

You can think of the three books who make a volume as a single big book (their relative positions can't change) and so for the other two volumes books, they can be considered as a single one too since their relative position can't change, therefore you have the equivalent of 5 books:

(3 books), (2 books), (1 book), (1 book), (1 book)

And the number of ways you can arrange them is $5!$ indeed.

Answer 2

Since the $3$-volume series are placed together according to their correct order, we have $A_{1}A_{2}A_{3}$ and for the $2$ volume series we also have $B_{1}B_{2}.$

So we can consider the $3$ volume books to be a single book denoted by $A$ and the $2$ volume books to be a single book denoted by $B$. Thus we need to arrange $A,B$ and the remaining $3$ books. This is the same as the number of ways to rearrange $5$ books.

Since the number of ways in which $5$ books can be arranged is $5!=120,$ we have $120$ ways.