Question:
In how many ways can 8 different chocolates be distributed among 5 children such that each can get any number of chocolates (including zero)?
In how many ways can 8 different chocolates be distributed among 5 children such that each can get any number of chocolates (including zero)?
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Always ask: "Who has the choice?" The chocolate can choose any child, but a child doesn't "choose" a specific number of chocolates in a fixed way. The items being distributed always form the power (exponent).
Updated On: Apr 20, 2026
- \(^{8}P_{5}\)
- \(^{8}C_{5}\)
- \(8^{5}\)
- \(5^{8}\)
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The Correct Option is D
Solution and Explanation
Step 1: Understanding the Concept:
Similar to the letter-box problem, each distinct chocolate "decides" which child it goes to. Since there are no restrictions on how many chocolates a child can receive, every chocolate has the same number of options.
Step 2: Detailed Explanation:
1. There are 8 different chocolates.
2. For the 1st chocolate, there are 5 choices (Child 1, 2, 3, 4, or 5).
3. For the 2nd chocolate, there are again 5 choices.
4. This continues until the 8th chocolate, which also has 5 choices.
5. Total number of ways = \( 5 \times 5 \times \dots \text{ (8 times)} = 5^8 \).
Step 3: Final Answer:
The chocolates can be distributed in \(5^{8}\) ways.
Similar to the letter-box problem, each distinct chocolate "decides" which child it goes to. Since there are no restrictions on how many chocolates a child can receive, every chocolate has the same number of options.
Step 2: Detailed Explanation:
1. There are 8 different chocolates.
2. For the 1st chocolate, there are 5 choices (Child 1, 2, 3, 4, or 5).
3. For the 2nd chocolate, there are again 5 choices.
4. This continues until the 8th chocolate, which also has 5 choices.
5. Total number of ways = \( 5 \times 5 \times \dots \text{ (8 times)} = 5^8 \).
Step 3: Final Answer:
The chocolates can be distributed in \(5^{8}\) ways.
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