Permutations & Combinations
Distribution of Objects
Grade 11
Question:
<p>The total number of ways in which 5 balls of different colours can be distributed among 3 persons so that each person gets at least one ball is</p>
<p>(A) 75</p>
<p>(B) 150</p>
<p>(C) 210</p>
<p>(D) 243</p>
Step-by-Step Solution
Key Concept: Use inclusion-exclusion principle: total distributions minus those missing at least one person.
<p><strong>Solution:</strong> Each of 5 distinct balls can go to any of 3 persons, giving \(3^5 = 243\) total ways.</p><p>Subtract cases where at least one person gets no ball using inclusion-exclusion:</p><p>- Ways where at least 1 person gets nothing: \(\binom{3}{1} \cdot 2^5 = 3 \cdot 32 = 96\)</p><p>- Ways where at least 2 persons get nothing: \(\binom{3}{2} \cdot 1^5 = 3 \cdot 1 = 3\)</p><p>- Ways where all 3 get nothing: 0 (impossible)</p><p>By inclusion-exclusion: \(243 - 96 + 3 = 150\)</p>
Correct Answer: B