In how many ways can 4 balls be placed in 4 boxes if 3 balls are indistinguishably white, 1 ball is black, and the boxes are indistinguishable?

RedDragon1 Oct 19, 2024

#1**-1 **

To solve the problem, we need to find how many distinct ways we can distribute 4 balls (3 indistinguishable white balls and 1 distinguishable black ball) into 4 indistinguishable boxes.

Let's denote the number of white balls in each box by \( a_1 \), \( a_2 \), \( a_3 \), and \( a_4 \), where \( a_i \) represents the number of white balls in box \( i \) and we need to find non-negative integer solutions to the equation:

\[

a_1 + a_2 + a_3 + a_4 = 3

\]

We also have to take into account the placement of the black ball after distributing the white balls.

### Step 1: Count the distributions of white balls

First, we will enumerate the different partitions of the number 3, corresponding to how we can allocate 3 indistinguishable white balls into indistinguishable boxes. The partitions of 3 are:

1. \( 3 \) (All three white balls in one box)

2. \( 2 + 1 \) (Two white balls in one box and one white ball in another)

3. \( 1 + 1 + 1 \) (One white ball in each of three boxes)

Now we can list these partitions in a way that clearly defines the placement in indistinguishable boxes:

- **Partition \( 3 \)**: (3)

- **Partition \( 2 + 1 \)**: (2, 1)

- **Partition \( 1 + 1 + 1 \)**: (1, 1, 1)

### Step 2: Place the black ball

Now we need to consider how the black ball can be placed in relation to each distribution of the white balls:

1. **For the distribution of \( 3 \)**:

- There is only 1 box with all 3 white balls. When we add the black ball, it can go into this box or one of the remaining (empty) boxes.

- Total configurations = 1 (black in the box) + 3 (black in empty boxes) = **1 way**.

2. **For the distribution of \( 2 + 1 \)**:

- We have one box with 2 white balls and one box with 1 white ball.

- Possible locations for the black ball:

- It can go into the box with 2 white balls (1 way)

- It can go into the box with 1 white ball (1 way)

- It can go into one of the empty boxes (2 empty boxes)

- Therefore, total configurations = 1 + 1 + 2 = **4 ways**.

3. **For the distribution of \( 1 + 1 + 1 \)**:

- There are 3 boxes, each containing 1 white ball.

- The black ball can go into any of these 3 boxes (there are no empty boxes since each has a white ball).

- Total configurations = 3 (the black ball can go into any of the three boxes) = **3 ways**.

### Step 3: Sum the possibilities

Finally, we sum all the possible placements derived from each distribution:

\[

1 + 4 + 3 = 8

\]

Thus, the total number of ways to place 4 balls (3 indistinguishably white and 1 distinguishable black) into 4 indistinguishable boxes is **8**.

Therefore, the answer is:

\[

\boxed{8}

\]

AUnVerifedTaxPayer Oct 19, 2024