Placing Books on Shelves Permutation Question Question: There are five distinct A, three distinct B
books, and two distinct C books. In how many ways can these
books be arranged on a shelf if one of the B books is on the
left of all A books, and another B book is on
the right of all the A books?
This is what I've done:
Step 1: Place all A books first so, 5!
Step 2: Place B books
3 choose 2 ways to place B books + 4 ways to place the remaining B book
Step 3: Place C books
5 choose 2 ways to place C books
Is this the correct way to approach this question? I got 24000 ways as an answer.
Thanks,

*

*Choose an order for the A books . . . . . $5!$ ways

*Choose the leftmost B book . . . . . $3$ ways

*Choose the rightmost B book . . . . . $2$ ways

*At this stage we have B A A A A A B.  Choose one of the spaces to take the last B book . . . . . $6$ ways.

*We have now a row of $8$ books; choose a place for the first (e.g., in alphabetical order) of the C books . . . . . $9$ ways.

*Choose a place for the other C book . . . . .$10$ ways.


Answer: $5!\times3\times2\times6\times10\times9=388800$.

Comment.  It is important to always be clear on what you are choosing.  For example, when you say "$5$ choose $2$ ways to place C books", this would mean you are choosing $2$ somethings from $5$ possible somethings.  I don't what are the "somethings" you are choosing.  What you should be choosing are two places for the books, and at this stage there are $9$ places available, not $5$.
