Find the number of ways in which 21 balls can be distributed among 3 persons such that each person does not receive less than 5 balls.

Options

A. 28
B. 14
C. 21
D. 7

Answer

To solve the problem of distributing 21 balls among 3 persons such that each person receives at least 5 balls, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Problem**: We need to distribute 21 balls to 3 persons (let's call them A, B, and C) with the condition that each person receives at least 5 balls.

2. **Initial Distribution**: Since each person must receive at least 5 balls, we start by giving 5 balls to each person. This means:
- A receives 5 balls
- B receives 5 balls
- C receives 5 balls

Total balls distributed so far = 5 + 5 + 5 = 15 balls.

3. **Calculating Remaining Balls**: After distributing 15 balls, we have:
\[ 21 - 15 = 6 \text{ balls remaining} \]

4. **Distributing Remaining Balls**: Now, we need to distribute the remaining 6 balls among the 3 persons (A, B, and C) without any restrictions (i.e., a person can receive 0 or more of the remaining balls).

5. **Using the Stars and Bars Theorem**: The problem of distributing \( n \) identical items (remaining balls) into \( r \) distinct groups (persons) can be solved using the "stars and bars" theorem. The formula is:
\[ \text{Number of ways} = \binom{n + r - 1}{r - 1} \]
Here, \( n = 6 \) (remaining balls) and \( r = 3 \) (persons).

6. **Applying the Formula**:
\[ \text{Number of ways} = \binom{6 + 3 - 1}{3 - 1} = \binom{8}{2} \]

7. **Calculating \( \binom{8}{2} \)**:
\[ \binom{8}{2} = \frac{8 \times 7}{2 \times 1} = \frac{56}{2} = 28 \]

8. **Final Answer**: Therefore, the number of ways to distribute the 21 balls among 3 persons such that each person receives at least 5 balls is **28**.