Permutations & Combinations
Group arrangements
Grade 11
Question:
<p>Three boys of class X, four boys of class XI, and five boys of class XII sit in a row. The total number of ways in which these boys can sit so that all the boys of same class sit together is equal to</p>
<p>\((3!)^2(4!)(5!)\)</p>
<p>\((3!)(4!)^2(5!)\)</p>
<p>\((3!)(4!)(5!)\)</p>
<p>\((3!)(4!)(5!)^2\)</p>
Step-by-Step Solution
Key Concept: Treat each class group as a single unit first, then arrange these 3 units among themselves, and finally arrange boys within each unit independently.
<p><strong>Step 1:</strong> Identify the constraint - all boys of the same class must sit together. This means we have 3 distinct groups (Class X, Class XI, Class XII).</p><p><strong>Step 2:</strong> Treat each class as a single unit/block. Now arrange these 3 blocks in a row: <strong>3! = 6 ways</strong></p><p><strong>Step 3:</strong> Within Class X block, arrange 3 boys: <strong>3! = 6 ways</strong></p><p><strong>Step 4:</strong> Within Class XI block, arrange 4 boys: <strong>4! = 24 ways</strong></p><p><strong>Step 5:</strong> Within Class XII block, arrange 5 boys: <strong>5! = 120 ways</strong></p><p><strong>Step 6:</strong> By multiplication principle, total arrangements = <strong>3! × 3! × 4! × 5! = 6 × 6 × 24 × 120 = <u>103,680</u></strong></p><p>∴ Answer: A</p>
Correct Answer: A