Number of ways to place 5 balls in 3 boxes

To determine the number of ways to place 5 identical balls in 3 identical boxes such that no box remains empty, we can use the concept of stars and bars.

Stars and Bars method

The stars and bars method is a combinatorial technique used to count the number of ways to distribute identical objects into distinct groups. In this case, the stars represent the balls, and the bars represent the dividers between the boxes.

Step 1: Placement of bars

Since we have 3 boxes, we need to place 2 bars to create 3 divisions. However, we cannot have consecutive bars as this would result in an empty box. Therefore, we have the following possibilities for placing the bars:

- *|*|* (One ball in each box)
- **|*| (Two balls in the first box, and one ball in the second box)
- *|**| (One ball in the first box, and two balls in the second box)

Step 2: Placement of balls

Now that we have placed the bars, we need to distribute the 5 balls among the boxes. Since the balls are identical and the boxes are identical, we need to consider the number of ways to distribute the balls among the divisions.

- *|*|*: There is only one way to distribute the 5 balls, which is to place 1 ball in each box.
- **|*|: There are 6 possible distributions of the 5 balls among the boxes: (3, 2, 0), (2, 3, 0), (1, 4, 0), (3, 0, 2), (2, 0, 3), (1, 0, 4). These represent the number of balls in each box.
- *|**|: Similarly, there are 6 possible distributions of the 5 balls among the boxes: (3, 0, 2), (2, 0, 3), (1, 0, 4), (3, 2, 0), (2, 3, 0), (1, 4, 0).

Step 3: Total number of ways

To determine the total number of ways to place the 5 balls in 3 boxes, we add up the possibilities from each case:

- *|*|*: 1 way
- **|*|: 6 ways
- *|**|: 6 ways

Therefore, the total number of ways is 1 + 6 + 6 = 13.

However, since we are asked to find the number of ways such that no box remains empty, we need to subtract the case where all the balls are in one box (3, 2, 0) or (0, 2, 3). Therefore, the final answer is 13 - 2 = 11.

Hence, the correct answer is 2.