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There are four different types of bananas. In how many ways can 12 children select bananas so that at least one child selects different types of bananas?

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For “at least one” type problems, it is often easier to count the total possibilities and subtract the unwanted cases using the complement principle.
Updated On: Feb 15, 2026
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Solution and Explanation

Step 1: Understanding the Topic
This is a problem in combinatorics that involves the "at least one" condition. Such problems are often best solved using the principle of complementary counting. We will calculate the total number of ways without any restrictions and then subtract the number of ways that violate the condition.
Step 2: Key Approach - Complementary Counting
The condition is "at least one child selects different types of bananas". The opposite (complement) of this condition is "all children select the exact same type of banana". The strategy is: \[ (\text{Required Ways}) = (\text{Total Ways}) - (\text{Ways all children choose the same type}) \] Step 3: Detailed Calculation
A. Calculate the total number of ways:
There are 12 distinct children, and each child can independently choose any one of the 4 types of bananas.

Child 1 has 4 choices.
Child 2 has 4 choices.
...
Child 12 has 4 choices.
By the multiplication principle, the total number of ways is: \[ \text{Total Ways} = 4 \times 4 \times \dots \times 4 \text{ (12 times)} = 4^{12} \] B. Calculate the number of ways for the complementary case:
We need to count the scenarios where all 12 children make the same choice.

Case 1: All 12 children choose banana type 1. (1 way)
Case 2: All 12 children choose banana type 2. (1 way)
Case 3: All 12 children choose banana type 3. (1 way)
Case 4: All 12 children choose banana type 4. (1 way)
The total number of ways for this complementary case is $1+1+1+1 = 4$.
C. Find the required number of ways:
\[ \text{Required Ways} = 4^{12} - 4 \] Step 4: Final Answer
The number of ways is $4^{12} - 4$. \[ \boxed{4^{12} - 4} \]
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