Key Concepts and Formulas

• Inclusion-Exclusion Principle: For sets $A_1, A_2, \dots, A_n$, $|A_1 \cup A_2 \cup \dots \cup A_n| = \sum_{i=1}^n |A_i| - \sum_{1 \le i < j \le n} |A_i \cap A_j| + \sum_{1 \le i < j < k \le n} |A_i \cap A_j \cap A_k| - \dots + (-1)^{n-1} |A_1 \cap A_2 \cap \dots \cap A_n|$
• Distributing distinct objects into distinct boxes: The number of ways to distribute $n$ distinct objects into $k$ distinct boxes is $k^n$.
• Complementary Counting: The number of ways to have an event occur is the total number of possibilities minus the number of ways the event does not occur.

Step-by-Step Solution

Step 1: Define the Universal Set and the Sets of Interest

We want to distribute 20 distinct oranges to 3 distinct children such that each child gets at least one orange. Let $U$ be the set of all possible ways to distribute the oranges without any restrictions. Let $A_i$ be the set of distributions where child $i$ receives no oranges, for $i = 1, 2, 3$. We want to find $|U| - |A_1 \cup A_2 \cup A_3|$.

Step 2: Calculate the Size of the Universal Set $|U|$

Each of the 20 distinct oranges can be given to any of the 3 children. Since the oranges are distinct, we have 3 choices for each orange. Therefore, the total number of ways to distribute the oranges without any restrictions is: $|U| = 3^{20}$

Step 3: Calculate the Size of the Union $|A_1 \cup A_2 \cup A_3|$ using the Inclusion-Exclusion Principle

We want to find the number of ways where at least one child receives no oranges. Using the Inclusion-Exclusion Principle: $|A_1 \cup A_2 \cup A_3| = |A_1| + |A_2| + |A_3| - |A_1 \cap A_2| - |A_1 \cap A_3| - |A_2 \cap A_3| + |A_1 \cap A_2 \cap A_3|$

Step 4: Calculate $|A_1| + |A_2| + |A_3|$

$|A_1|$ is the number of ways where child 1 receives no oranges. This means each of the 20 oranges can only go to child 2 or child 3. So, $|A_1| = 2^{20}$. Similarly, $|A_2| = 2^{20}$ and $|A_3| = 2^{20}$. Therefore, $|A_1| + |A_2| + |A_3| = 3 \cdot 2^{20}$

Step 5: Calculate $|A_1 \cap A_2| + |A_1 \cap A_3| + |A_2 \cap A_3|$

$|A_1 \cap A_2|$ is the number of ways where child 1 and child 2 receive no oranges. This means each of the 20 oranges must go to child 3. So, $|A_1 \cap A_2| = 1^{20} = 1$. Similarly, $|A_1 \cap A_3| = 1$ and $|A_2 \cap A_3| = 1$. Therefore, $|A_1 \cap A_2| + |A_1 \cap A_3| + |A_2 \cap A_3| = 3 \cdot 1 = 3$

Step 6: Calculate $|A_1 \cap A_2 \cap A_3|$

$|A_1 \cap A_2 \cap A_3|$ is the number of ways where all three children receive no oranges. Since we have 20 oranges to distribute, this is impossible. Thus, $|A_1 \cap A_2 \cap A_3| = 0$.

Step 7: Substitute the Values into the Inclusion-Exclusion Formula

$|A_1 \cup A_2 \cup A_3| = 3 \cdot 2^{20} - 3 + 0 = 3 \cdot 2^{20} - 3$

Step 8: Calculate the Final Answer

The number of ways to distribute the oranges such that each child gets at least one orange is: $|U| - |A_1 \cup A_2 \cup A_3| = 3^{20} - (3 \cdot 2^{20} - 3) = 3^{20} - 3 \cdot 2^{20} + 3$

Common Mistakes & Tips

• Remember to use the Inclusion-Exclusion Principle when dealing with "at least" or "at most" type problems.
• Carefully consider whether the objects and recipients are distinct or identical. In this case, both are distinct.
• Don't forget the alternating signs in the Inclusion-Exclusion formula.

Summary

We used the Inclusion-Exclusion Principle to find the number of ways to distribute 20 distinct oranges to 3 distinct children such that each child receives at least one orange. We first calculated the total number of ways to distribute the oranges without any restrictions and then subtracted the number of ways where at least one child receives no orange. This gave us the final answer: $3^{20} - 3 \cdot 2^{20} + 3$.

The final answer is \boxed{3^{20} - 3 \cdot 2^{20} + 3}. This corresponds to option (3).