Permutations & Combinations
Distribution with Parity Constraint
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 an even number of white balls.</p>
<p>(A) 844</p>
<p>(B) 929</p>
<p>(C) 948</p>
<p>(D) 1017</p>
Step-by-Step Solution
Key Concept: Count distributions where the white ball count is restricted to even values using summation of binomial coefficients.
<p><strong>Step 1:</strong> Let the number of white balls be \(w\) (even). Then \(r + g + b = 20 - w\).</p><p><strong>Step 2:</strong> For even \(w\): \(w = 0, 2, 4, ..., 20\).</p><p><strong>Step 3:</strong> For each even value of \(w\), the number of ways to distribute the remaining \(20-w\) balls among red, green, and blue is \(\binom{20-w+2}{2}\).</p><p><strong>Step 4:</strong> Sum over all even values:</p><p>\[\sum_{k=0}^{10} \binom{22-2k}{2} = \binom{22}{2} + \binom{20}{2} + \binom{18}{2} + ... + \binom{2}{2}\]</p><p><strong>Step 5:</strong> \[= 231 + 190 + 153 + 120 + 91 + 66 + 45 + 28 + 15 + 6 + 1 = 948\]</p><p>∴ Answer is <strong>(C) 948</strong></p>
Correct Answer: C