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

We need to find the number of ways of arranging 8 identical books into 4 identical shelves, where any number of shelves may remain empty.

Since both the books and the shelves are identical, this problem reduces to finding the number of partitions of 8 into at most 4 parts.

A partition of 8 into at most 4 parts means writing 8 as an unordered sum of at most 4 positive integers (where empty shelves correspond to parts of value 0).

Listing all partitions of 8 into at most 4 parts:

1 part:

8 = 8 → (8)

2 parts:

8 = 7+1, 6+2, 5+3, 4+4 → 4 partitions

3 parts:

8 = 6+1+1, 5+2+1, 4+3+1, 4+2+2, 3+3+2 → 5 partitions

4 parts:

8 = 5+1+1+1, 4+2+1+1, 3+3+1+1, 3+2+2+1, 2+2+2+2 → 5 partitions

Total = 1 + 4 + 5 + 5 = 15

Therefore, the correct answer is Option 4: 15.