<p>A shelf contains 20 books of which 4 are single volume and the other form sets of 8, 5, and 3 volumes. Find the number of ways in which the books may be arranged on the shelf so that<br>(i) volumes of each set will not be separated,<br>(ii) volumes of each set remain in their due order.</p>
Step-by-Step Solution
Key Concept: Treat each complete set as a single unit for arrangement, then account for internal arrangements within sets. For part (ii), fixing the internal order means no permutation within each set—only arrange the units themselves.
<p><strong>Part (i): Volumes of each set will NOT be separated</strong></p><p><strong>Step 1:</strong> Identify the units to arrange: 4 single books + 1 set of 8 volumes + 1 set of 5 volumes + 1 set of 3 volumes = 7 distinct units total.</p><p><strong>Step 2:</strong> Arrange these 7 units on the shelf: <strong>7!</strong> ways.</p><p><strong>Step 3:</strong> Within the set of 8 volumes, arrange them among themselves: <strong>8!</strong> ways.</p><p><strong>Step 4:</strong> Within the set of 5 volumes, arrange them: <strong>5!</strong> ways.</p><p><strong>Step 5:</strong> Within the set of 3 volumes, arrange them: <strong>3!</strong> ways.</p><p><strong>Total for (i):</strong> <strong>7! × 8! × 5! × 3!</strong></p><p><strong>Part (ii): Volumes of each set remain in their DUE ORDER</strong></p><p><strong>Step 1:</strong> 'Due order' means the internal sequence within each set is FIXED—no permutation allowed within sets.</p><p><strong>Step 2:</strong> We only arrange the 7 units (4 single books + 3 sets as fixed blocks): <strong>7!</strong> ways.</p><p><strong>Step 3:</strong> Since the order within each set is predetermined, we select positions for 20 books where 8 go to set 1, 5 to set 2, 3 to set 3, and 4 are singles: <strong>$\binom{20}{8,5,3,4} = \frac{20!}{8! \times 5! \times 3! \times 4!}$</strong></p><p><strong>Alternatively:</strong> Arrange 7 units in 7! ways, then arrange within as fixed sequences = <strong>$\frac{20!}{8! \times 5! \times 3!}$</strong> ÷ 4! if singles were identical, but they're distinct, so answer is <strong>$\frac{20!}{8! \times 5! \times 3!}$</strong></p><p>∴ <strong>Answer: (i) 7! × 8! × 5! × 3! (ii) $\frac{20!}{8! \times 5! \times 3!}$</strong></p>
Correct Answer: (i) \(7! \times 8! \times 5! \times 3!\), (ii) \(\frac{20!}{8! \times 5! \times 3!} \times \frac{1}{1}\)