Permutations & Combinations
Selection with repetition and constraints
Grade 11
Question:
<p><strong>238.</strong> 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>(a) 1001</p>
<p>(b) 286</p>
<p>(c) 680</p>
<p>(d) 455</p>
Step-by-Step Solution
Key Concept: Transform the constraint 'at least 2 of each colour' into an unconstrained problem by first removing 2 balls of each colour (8 balls total), then distribute the remaining 10 balls freely among 4 colours using stars and bars.
<p><strong>Step 1:</strong> Identify the constraint. We need to select 18 balls with at least 2 of each of the 4 colours (red, white, blue, green).</p><p><strong>Step 2:</strong> Remove the minimum requirement. Since we must have at least 2 of each colour, first allocate 2 balls to each colour: 2 × 4 = 8 balls reserved.</p><p><strong>Step 3:</strong> Reduce the problem. We now need to distribute the remaining 18 - 8 = 10 balls freely among 4 colours with no restrictions.</p><p><strong>Step 4:</strong> Apply stars and bars formula. The number of ways to distribute n identical objects into r groups is C(n+r-1, r-1).</p><p>Here, n = 10 and r = 4, so the number of selections is:</p><p>k = C(10+4-1, 4-1) = C(13, 3) = (13 × 12 × 11)/(3 × 2 × 1) = 1716/6 = 286</p><p><strong>∴ Answer: D (k = 286)</strong></p>
Correct Answer: D