<p>A man has three friends. The number of ways he can invite one friend everyday for dinner on six successive nights so that no friend is invited more than three times is</p>
Step-by-Step Solution
Key Concept: Use inclusion-exclusion principle by counting total ways to invite 3 friends over 6 nights, then subtract cases where at least one friend is invited more than 3 times (which is impossible here since 6 nights total). Instead, directly count distributions where each friend gets 0-3 invitations and the total is exactly 6.
<p><strong>Step 1:</strong> Find all valid distributions (a,b,c) where a+b+c=6 and 0≤a,b,c≤3.</p><p>Valid distributions:</p><ul><li>(3,3,0) and permutations: 3 ways</li><li>(3,2,1) and permutations: 6 ways</li></ul><p><strong>Step 2:</strong> For each distribution, count ordered arrangements:</p><ul><li>Type (3,3,0): Choose positions for friend 1 to appear 3 times, friend 2 to appear 3 times = 6!/(3!3!) = 20 ways per permutation = 3×20 = 60</li><li>Type (3,2,1): Arrange 3,2,1 invitations = 6!/(3!2!1!) = 60 ways per permutation = 6×60 = 360</li></ul><p><strong>Step 3:</strong> Total = 60 + 360 = 420</p><p>∴ Answer: C (420)</p>
Correct Answer: C