<p>How many different committees of 5 members can be formed from 6 men and 4 ladies if each committee is to contain at least one lady?</p>
Step-by-Step Solution
Key Concept: Use complementary counting: subtract committees with NO ladies from total committees. This avoids tedious case-by-case enumeration (1 lady, 2 ladies, etc.).
<p><strong>Step 1:</strong> Find total committees with 5 members from 10 people (6 men + 4 ladies).</p><p>Total = C(10,5) = 252</p><p><strong>Step 2:</strong> Find committees with NO ladies (all 5 from 6 men only).</p><p>Committees with 0 ladies = C(6,5) = 6</p><p><strong>Step 3:</strong> Apply complementary counting principle.</p><p>Committees with at least 1 lady = Total - (Committees with 0 ladies)</p><p>= 252 - 6 = 246</p><p><strong>Verification:</strong> Direct method: C(6,4)C(4,1) + C(6,3)C(4,2) + C(6,2)C(4,3) + C(6,1)C(4,4) = 15(4) + 20(6) + 15(4) + 6(1) = 60 + 120 + 60 + 6 = 246 ✓</p><p>∴ Answer: <strong>246</strong></p>
Correct Answer: 246