Permutations & Combinations
Distribution with Constraints
Grade 11
Question:
<p>n different toys have to be distributed among n children. Total number of ways in which these toys can be distributed so that exactly one child gets no toy, is equal to</p>
<p>(a) \(n!\)</p>
<p>(b) \(n! \cdot \binom{n}{2}\)</p>
<p>(c) \((n-1)! \cdot \binom{n}{2}\)</p>
<p>(d) \(n! \cdot \binom{n-1}{2}\)</p>
Step-by-Step Solution
Key Concept: Choose the child who receives nothing, then ensure the remaining toys are distributed such that exactly one child receives nothing.
<p>First choose which child gets no toy: $\binom{n}{1} = n$ ways. Then distribute n toys among the remaining $(n-1)$ children such that each gets at least one toy. By inclusion-exclusion, this is $(n-1)! \cdot \binom{n}{2}$ (considering surjective distributions and overcounting).</p><p>Alternatively: Choose 2 children to share toys such that one gets none. The count is $n! \cdot \binom{n}{2}$.</p>
Correct Answer: B