3. How many ways are there to line up 4 type A books, 2 type B books, and 3 type C books, given that all type A books are indistinguishable, each type B book is somehow identical to 3 type A books, and is distinguishable from the remaining one (given that each type B book is distinguishable with a different type A book), and all type C books are distinguishable??
To calculate the number of ways to line up the books, we can use the concept of permutations.
First, let's consider the type A books. Since they are indistinguishable, we can treat them as a single unit. So, we have 1 group of A books, 2 B books, and 3 C books.
To arrange these groups, we can use permutations.
Step 1: Permute the different groups.
There are 3 groups: A books, B books, and C books. We arrange them in all possible orders: 3!
Step 2: Permute the books within each group.
For the A books, they are indistinguishable, so we don't need to permute them.
For the B books, we can choose 1 out of 2 books to be placed first. Then, we permute the remaining book with the remaining 3 A books. So, we have 2 options for the first B book, and 4! options for the remaining 4 books (3 A books + 1 B book).
For the C books, all 3 are distinguishable, so we can arrange them in 3! ways.
Step 3: Multiply the results from each step to get the final answer.
Finally, we multiply the results from each step to get the total number of ways to line up the books:
3! * (2 * 4!) * 3!
Simplifying this expression, we get:
6 * (2 * 24) * 6 = 6 * 48 * 6 = 1,728
Therefore, there are 1,728 ways to line up the books.