Solution: We need to count the ways of arranging 8 identical books into 4 identical shelves, allowing any number of shelves to be empty.
Step 1. 3 Shelves Empty: Only one shelf holds all 8 books:
\((8, 0, 0, 0) \implies 1 \text{ way}\)
Step 2. 2 Shelves Empty: Two shelves hold the books in pairs of configurations:
\((7, 1, 0, 0), (6, 2, 0, 0), (5, 3, 0, 0), (4, 4, 0, 0) \implies 4 \text{ ways}\)
Step 3. 1 Shelf Empty: Three shelves hold the books in possible configurations:
\((6, 1, 1, 0), (5, 2, 1, 0), (4, 3, 1, 0), (4, 2, 2, 0), (3, 3, 2, 0) \implies 5 \text{ ways}\)
Step 4. 0 Shelves Empty: All four shelves hold the books in possible configurations:
\((5, 1, 1, 1), (4, 2, 1, 1), (3, 3, 1, 1), (3, 2, 2, 1), (2, 2, 2, 2) \implies 5 \text{ ways}\)
Adding up all the ways:
\(\text{Total} = 1 + 4 + 5 + 5 = 15 \text{ ways}\)
The Correct Answer is: 15 ways
How many possible words can be created from the letters R, A, N, D (with repetition)?
Let A be a 3 × 3 matrix such that \(\text{det}(A) = 5\). If \(\text{det}(3 \, \text{adj}(2A)) = 2^{\alpha \cdot 3^{\beta} \cdot 5^{\gamma}}\), then \( (\alpha + \beta + \gamma) \) is equal to: