Permutations & Combinations
Selection with repetition
Grade 11
Question:
<p>From an unlimited number of red, white, blue and green balls, a selection of 18 balls is to be made so that there are at least two of each colour. If the number of selection is \(k\), then \(k\) is equal to:</p>
<p>1001</p>
<p>286</p>
<p>680</p>
<p>455</p>
Step-by-Step Solution
Key Concept: Transform the constraint 'at least 2 of each colour' into an unrestricted problem by first allocating 2 balls to each colour (using 8 balls), then distributing the remaining 10 balls freely among 4 colours.
<p><strong>Step 1:</strong> Identify the constraint. We need to select 18 balls with at least 2 balls of each of the 4 colours (red, white, blue, green).</p><p><strong>Step 2:</strong> Apply constraint transformation. Allocate 2 balls to each colour first: 2×4 = 8 balls used. Remaining balls to distribute: 18 - 8 = 10 balls.</p><p><strong>Step 3:</strong> Convert to unrestricted problem. Now we need to distribute 10 identical balls among 4 distinct colours with no restrictions. This is equivalent to finding non-negative integer solutions to: r + w + b + g = 10, where r, w, b, g represent additional balls beyond the mandatory 2.</p><p><strong>Step 4:</strong> Apply stars and bars formula. The number of ways = C(n + k - 1, k - 1) where n = 10 (balls to distribute) and k = 4 (colours).</p><p>Number of selections = C(10 + 4 - 1, 4 - 1) = C(13, 3) = (13 × 12 × 11)/(3 × 2 × 1) = 1716/6 = 286</p><p>∴ k = 286</p>
Correct Answer: D