In how many ways 10 green balls of same size can be placed in 3 distinct boxes, if no box remains empty?

Answer

To solve the problem of distributing 10 green balls into 3 distinct boxes such that no box remains empty, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Problem**: We need to distribute 10 identical green balls into 3 distinct boxes with the condition that no box can be empty.

2. **Initial Distribution**: Since no box can be empty, we must place at least one ball in each box. Therefore, we start by placing 1 ball in each of the 3 boxes. This uses up 3 balls, leaving us with:
\[ 10 - 3 = 7 \text{ balls remaining} \]

3. **Reformulate the Problem**: Now, we need to distribute the remaining 7 balls into the 3 boxes. There are no restrictions now since each box already has one ball.

4. **Apply the Stars and Bars Theorem**: The problem of distributing \( n \) identical items (balls) into \( r \) distinct groups (boxes) can be solved using the "Stars and Bars" theorem. The formula for this is given by:
\[ \binom{n + r - 1}{r - 1} \]
In our case, \( n = 7 \) (the remaining balls) and \( r = 3 \) (the boxes).

5. **Calculate the Combinations**: Plugging in the values into the formula:
\[ \binom{7 + 3 - 1}{3 - 1} = \binom{9}{2} \]

6. **Compute \( \binom{9}{2} \)**:
\[ \binom{9}{2} = \frac{9 \times 8}{2 \times 1} = \frac{72}{2} = 36 \]

7. **Final Answer**: Therefore, the total number of ways to distribute the 10 green balls into 3 distinct boxes such that no box remains empty is:
\[ \boxed{36} \]