Permutations & Combinations
Selection with restrictions
Grade 11
Question:
<p>To fill 12 vacancies there are 25 candidates of which 5 are from scheduled caste. If three of the vacancies are reserved for scheduled caste candidates while the rest are open to all, the number of ways in which the selection can be made is</p>
<p>\({}^5C_3 \times {}^{22}C_9\)</p>
<p>\({}^{22}C_9 - {}^5C_3\)</p>
<p>\({}^{22}C_3 + {}^5C_3\)</p>
<p>none of these</p>
Step-by-Step Solution
Key Concept: Partition the selection into two independent groups: first fill 3 reserved SC positions from 5 SC candidates, then fill remaining 9 open positions from all 25 candidates. Multiply these counts since they're independent events.
<p><strong>Step 1:</strong> Identify the constraint structure: 3 vacancies reserved for SC candidates, 9 vacancies open to all 25 candidates.</p><p><strong>Step 2:</strong> Fill reserved SC positions: Select 3 SC candidates from 5 available SC candidates = C(5,3) = 10 ways</p><p><strong>Step 3:</strong> Fill open positions: Select 9 candidates from all 25 candidates (SC candidates who weren't selected for reserved positions are still eligible) = C(25,9) ways</p><p><strong>Step 4:</strong> Apply multiplication principle since selections are independent: Total ways = C(5,3) × C(25,9) = 10 × C(25,9)</p><p>∴ Answer: A</p>
Correct Answer: A