Problem 3:
To solve this problem, we need to consider how distinguishable balls can be assigned to indistinguishable boxes. The key here is to count the distinct distributions of balls among the boxes without regard for which box is which.
Let's denote the balls as \(A\), \(B\), \(C\), and \(D\). Since the boxes are indistinguishable, we can only represent the counts of balls in each box. We can categorize the possible distributions based on how many boxes are used:
1. **All 4 balls in 1 box:**
- There is only one way to do this: \((4)\).
2. **3 balls in one box and 1 ball in another:**
- We can choose 1 ball to be separate. The number of ways to choose 1 ball out of 4 is \( \binom{4}{1} = 4 \).
- This distribution corresponds to \((3, 1)\).
3. **2 balls in one box and 2 balls in another:**
- We need to choose 2 balls to be together. The number of ways to choose 2 balls out of 4 is \( \binom{4}{2} = 6 \).
- This distribution corresponds to \((2, 2)\).
4. **2 balls in one box, 1 ball in a second box, and 1 ball in a third box:**
- First, we choose 2 balls to go in the first box, and the remaining two will each go in their own box. The number of ways to choose which 2 out of 4 balls go together is \( \binom{4}{2} = 6 \).
- This distribution corresponds to \((2, 1, 1)\).
Now, we can summarize the distributions:
- One way to have all balls in one box: \( (4) \)
- Four ways to have one box with three balls and another with one ball: \( (3, 1) \)
- Six ways to have two boxes with two balls in each: \( (2, 2) \)
- Six ways to have two boxes with one ball each and one box with two balls: \( (2, 1, 1) \)
Now we can tally these distinct distributions:
- 1 way for \( (4) \)
- 4 ways for \( (3, 1) \)
- 6 ways for \( (2, 2) \)
- 6 ways for \( (2, 1, 1) \)
To find the total number of ways to place the balls in the boxes, we add these up:
\[
1 + 4 + 6 + 6 = 17
\]
Therefore, the total number of ways to place 4 distinguishable balls into 3 indistinguishable boxes is \(\boxed{17}\).
