Number of ways of arranging 8 identical books into 4 identical shelves where any number of shelves may remain empty is equal to

Options

A. 18
B. 16
C. 12
D. 15

Answer

To solve the problem of arranging 8 identical books into 4 identical shelves where any number of shelves may remain empty, we can use the "stars and bars" theorem, which is a common combinatorial method for distributing indistinguishable objects (books) into distinguishable boxes (shelves). However, since the shelves are identical, we will need to consider the partitions of the number of books.

### Step-by-step Solution:

1. **Understanding the Problem**: We need to find the number of ways to distribute 8 identical books into 4 identical shelves. The shelves can be empty, which means we can have any number of books in each shelf, including zero.

2. **Using Partitions**: Since the shelves are identical, we need to find the partitions of the number 8 into at most 4 parts. Each part represents the number of books in a shelf.

3. **Finding Partitions**: We list the possible partitions of 8 into up to 4 parts:
- (8)
- (7, 1)
- (6, 2)
- (6, 1, 1)
- (5, 3)
- (5, 2, 1)
- (5, 1, 1, 1)
- (4, 4)
- (4, 3, 1)
- (4, 2, 2)
- (4, 2, 1, 1)
- (4, 1, 1, 1, 1)
- (3, 3, 2)
- (3, 3, 1, 1)
- (3, 2, 2, 1)
- (3, 2, 1, 1, 1)
- (2, 2, 2, 2)

4. **Counting the Partitions**: We count the valid partitions listed above:
- (8) → 1 way
- (7, 1) → 1 way
- (6, 2) → 1 way
- (6, 1, 1) → 1 way
- (5, 3) → 1 way
- (5, 2, 1) → 1 way
- (5, 1, 1, 1) → 1 way
- (4, 4) → 1 way
- (4, 3, 1) → 1 way
- (4, 2, 2) → 1 way
- (4, 2, 1, 1) → 1 way
- (3, 3, 2) → 1 way
- (3, 3, 1, 1) → 1 way
- (3, 2, 2, 1) → 1 way
- (2, 2, 2, 2) → 1 way

5. **Total Count**: Adding these up gives us a total of 15 distinct ways to arrange the books into the shelves.

### Final Answer:
The number of ways of arranging 8 identical books into 4 identical shelves where any number of shelves may remain empty is **15**.