<p>A library has \(a\) copies of one book, \(b\) copies each of two books, \(c\) copies each of three books, and single copy of \(d\) books. Find the total number of ways in which these books can be arranged in a shelf.</p>
Step-by-Step Solution
Key Concept: This is a multinomial permutation problem where identical items within each category must be accounted for using the multinomial coefficient formula: total arrangements = (total items)! divided by the product of factorials of identical item counts.
<p><strong>Step 1:</strong> Count total number of books on the shelf.</p><p>Total books = a (from first type) + 2b (from second type) + 3c (from third type) + d (distinct books) = a + 2b + 3c + d</p><p><strong>Step 2:</strong> Identify which books are identical within their categories.</p><p>• First book type: a identical copies</p><p>• Second book type: b identical copies (one book), and b identical copies (another book) → contributes b! and b! to denominator</p><p>• Third book type: c identical copies (one book), c identical copies (second book), and c identical copies (third book) → contributes c!, c!, and c! to denominator</p><p>• Fourth book type: d books, each distinct (single copy each) → no repetition factor</p><p><strong>Step 3:</strong> Apply the multinomial permutation formula.</p><p>When arranging n objects where there are n₁ identical objects of type 1, n₂ of type 2, etc., the number of arrangements is:</p><p>$$\text{Arrangements} = \dfrac{n!}{n_1! \cdot n_2! \cdot \ldots \cdot n_k!}$$</p><p><strong>Step 4:</strong> Substitute our values.</p><p>$$\text{Total arrangements} = \dfrac{(a+2b+3c+d)!}{a! \cdot b! \cdot b! \cdot c! \cdot c! \cdot c! \cdot 1! \cdot 1! \cdots 1!}$$</p><p>Since 1! = 1 for the d distinct books:</p><p>∴ Answer: $$\dfrac{(a+2b+3c+d)!}{a!\,b!\,b!\,c!\,c!\,c!}$$</p>
Correct Answer: \(\dfrac{(a+2b+3c+d)!}{a!\,b!\,b!\,c!\,c!\,c!}\)