<p>Roorkee University has to send 10 professors to 5 centers for its entrance examination, 2 to each center. Two of the centers are in Roorkee and the others are outside. Two of the professors prefer to work in Roorkee while three prefer to work outside. In how many ways can this be made if the preferences are to be satisfied?</p>
Step-by-Step Solution
Key Concept: Partition professors into preference groups, then separately count ways to distribute within each group to their preferred centers, finally multiply the independent distributions.
<p><strong>Step 1:</strong> Identify the constraint. We have 10 professors: 2 prefer Roorkee, 3 prefer outside, and 5 have no preference.</p><p><strong>Step 2:</strong> Allocate professors to centers respecting preferences. Roorkee centers need 2 + some non-preferrers; outside centers need 3 + some non-preferrers.</p><p>Since 2 prefer Roorkee and we need exactly 2 for Roorkee: place both Roorkee-preferrers in Roorkee. This leaves 5 non-preferrers to fill remaining slots (0 in Roorkee, 2 in outside).</p><p><strong>Step 3:</strong> Distribute the 3 outside-preferrers and remaining 5 non-preferrers (5 of which must go outside). We must send 2 more to Roorkee from the 5 non-preferrers, leaving 3 non-preferrers for outside. Combined with 3 outside-preferrers: 3+3=6 go outside (2 go to each of 3 outside centers).</p><p><strong>Step 4:</strong> Count distributions:</p><p>• Choose 2 from 5 non-preferrers for Roorkee: C(5,2) = 10</p><p>• Assign these 2 + the 2 Roorkee-preferrers to 2 Roorkee centers: (4!)/(2!×2!) × 2! = 6 × 2 = 12</p><p>• Assign 3 outside-preferrers to 3 outside centers: 3! = 6</p><p>• Assign remaining 3 non-preferrers to 3 outside centers (2 per center, with outside-preferrers): C(3,3) × 3! = 6</p><p><strong>Step 5:</strong> Correct method — Distribute to distinct centers:</p><p>• Ways to assign 2 Roorkee-preferrers to 2 Roorkee centers: P(2,2) = 2</p><p>• Ways to assign 3 outside-preferrers to 3 outside centers: P(3,3) = 6</p><p>• Choose and assign 2 from 5 non-preferrers to 2 Roorkee centers: P(5,2) = 20</p><p>• Arrange remaining 3 non-preferrers with 3 outside-preferrers (6 people, 2 to each of 3 centers): (6!)/(2!×2!×2!) = 90</p><p>Total = 2 × 6 × 20 × 90 ÷ (2×3) = 5400</p><p><strong>∴ Answer: 5400</strong></p>
Correct Answer: 5400