<p>There are <em>n</em> different books and each book has <em>p</em> copies. The number of selection of books from these is</p>
<p>(a) \((p+1)^n - 1\)</p>
<p>(b) \((p+1)^n - 2\)</p>
<p>(c) \(n(p+1) - 1\)</p>
<p>(d) none of these</p>
Step-by-Step Solution
Key Concept: Each distinct book can be selected in (p+1) ways (0 copies, 1 copy, 2 copies, ..., p copies), and selections from different books are independent. Multiply the choices for all n books and subtract 1 for the null selection.
<p><strong>Step 1:</strong> For each of the n different books, we can select 0, 1, 2, ..., or p copies.</p><p><strong>Step 2:</strong> This gives (p+1) choices for each book independently.</p><p><strong>Step 3:</strong> By multiplication principle, total selections = (p+1)×(p+1)×...×(p+1) [n times] = (p+1)ⁿ</p><p><strong>Step 4:</strong> Subtract 1 to exclude the empty selection (selecting 0 books of all types).</p><p><strong>Step 5:</strong> Number of ways = (p+1)ⁿ - 1</p><p>∴ Answer: A</p>
Correct Answer: A