Permutations & Combinations
Distribution with lower bound
Grade 11

Question:

<p>Number of ways in which ₹18 can be distributed among four persons such that nobody receives less than ₹3 is</p>
<p>\(4^2\)</p>
<p>\(2^4\)</p>
<p>\(4!\)</p>
<p>84</p>

Step-by-Step Solution

Key Concept: Transform the constrained distribution problem by giving each person their minimum amount (₹3) first, then distribute the remaining amount (₹18 - 12 = ₹6) freely among four persons using stars and bars.
<p><strong>Step 1:</strong> Identify the constraint. Each of 4 persons must receive at least ₹3.</p><p><strong>Step 2:</strong> Give each person their minimum amount first: 4 × 3 = ₹12. This leaves ₹18 - ₹12 = ₹6 to distribute freely.</p><p><strong>Step 3:</strong> Now find the number of non-negative integer solutions to x₁ + x₂ + x₃ + x₄ = 6, where xᵢ represents additional money for person i.</p><p><strong>Step 4:</strong> Apply stars and bars formula: C(n + k - 1, k - 1) = C(6 + 4 - 1, 4 - 1) = C(9, 3)</p><p><strong>Step 5:</strong> Calculate: C(9, 3) = 9!/(3! × 6!) = (9 × 8 × 7)/(3 × 2 × 1) = 504/6 = 84</p><p>∴ Answer: <strong>D (84)</strong></p>
Correct Answer: D

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