Permutations & Combinations
Distribution of identical objects
Grade 11
Question:
<p>Since at least one ball is in each box, for no box empty the formula is \({}^{(n-1)}C_{(r-1)} = {}^{(10-1)}C_{(4-1)} = {}^9C_3\), where \(n\) is the number of identical balls and \(r\) is the number of distinct boxes. The number of ways of choosing any 3 places from 9 different places is \({}^9C_3\). The number of ways of distributing 10 identical balls into 4 distinct boxes such that each box has at least one ball is:</p>
<p>\({}^9C_2\)</p>
<p>\({}^9C_3\)</p>
<p>\({}^{10}C_3\)</p>
<p>\({}^{10}C_4\)</p>
Step-by-Step Solution
Key Concept: When distributing n identical objects into r distinct boxes with each box getting at least one object, first place one ball in each box (using r balls), then distribute the remaining (n-r) balls freely among r boxes using stars and bars: C(n-1, r-1).
<p><strong>Step 1:</strong> Identify the constraint. We have n=10 identical balls and r=4 distinct boxes, with each box requiring at least one ball.</p><p><strong>Step 2:</strong> Use the restricted distribution formula. When each of r boxes must contain at least one ball from n identical balls, the number of ways is $^{(n-1)}C_{(r-1)}$.</p><p><strong>Step 3:</strong> This formula works because: Reserve 1 ball for each of the 4 boxes (using 4 balls). We have 10-4=6 balls remaining. These 6 balls can be distributed freely among 4 boxes with no restrictions.</p><p><strong>Step 4:</strong> Distributing 6 identical balls into 4 distinct boxes = $^{(6+4-1)}C_{(4-1)} = ^9C_3$, which matches our formula $^{(10-1)}C_{(4-1)} = ^9C_3$.</p><p><strong>Step 5:</strong> Calculate: $^9C_3 = \frac{9!}{3! \cdot 6!} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = \frac{504}{6} = 84$</p><p>∴ Answer: <strong>B (84 ways)</strong></p>
Correct Answer: B