Permutations & Combinations
Distribution of Identical Objects
Grade 11
Question:
<p>A large pile contains red, white, green and blue balls (all alike except for colour). Find the number of ways to select 20 balls from them such that the selection has at most 3 green balls.</p>
<p>(A) 802</p>
<p>(B) 969</p>
<p>(C) 973</p>
<p>(D) 879</p>
Step-by-Step Solution
Key Concept: Use the complement principle with stars and bars to count distributions where green balls are constrained to at most 3.
<p><strong>Step 1:</strong> Let the number of red, white, green, and blue balls selected be \(r\), \(w\), \(g\), and \(b\) respectively.</p><p><strong>Step 2:</strong> We need \(r + w + g + b = 20\) with \(0 \leq g \leq 3\).</p><p><strong>Step 3:</strong> Using stars and bars, the number of non-negative integer solutions is:</p><p>\[\text{Total} = \sum_{g=0}^{3} \binom{20-g+3-1}{3-1} = \sum_{g=0}^{3} \binom{22-g}{2}\]</p><p><strong>Step 4:</strong> Calculate each term:</p><p>\[\binom{22}{2} + \binom{21}{2} + \binom{20}{2} + \binom{19}{2} = 231 + 210 + 190 + 171 = 802\]</p><p>∴ Answer is <strong>(A) 802</strong></p>
Correct Answer: C