Find the number of ways of distributing 7 identical balls into three boxes so that no box is empty and each box being large enough to accomodate all balls.

Options

A. 15
B. `7^3`
C. `""^(24)C_4`
D. `(20!)/(4!)`

Answer

To solve the problem of distributing 7 identical balls into 3 boxes such that no box is empty, we can follow these steps:

### Step 1: Understand the Problem
We have 7 identical balls and 3 boxes. The requirement is that no box should be empty.

### Step 2: Initial Distribution
Since no box can be empty, we need to place at least one ball in each box. Therefore, we will first place 1 ball in each of the 3 boxes.

- After placing 1 ball in each box, we have used 3 balls (1 in each of the 3 boxes).
- Remaining balls = 7 - 3 = 4 balls.

### Step 3: Use the Stars and Bars Theorem
Now, we need to distribute the remaining 4 balls into the 3 boxes. Since the boxes can now be empty (as we have already ensured that each box has at least one ball), we can use the "Stars and Bars" theorem.

The formula for distributing \( n \) identical items into \( r \) distinct groups (where groups can be empty) is given by:
\[ \text{Number of ways} = \binom{n + r - 1}{r - 1} \]
In our case:
- \( n = 4 \) (remaining balls)
- \( r = 3 \) (boxes)

### Step 4: Substitute Values into the Formula
Substituting the values into the formula:
\[ \text{Number of ways} = \binom{4 + 3 - 1}{3 - 1} = \binom{6}{2} \]

### Step 5: Calculate the Binomial Coefficient
Now we calculate \( \binom{6}{2} \):
\[ \binom{6}{2} = \frac{6!}{2!(6-2)!} = \frac{6 \times 5}{2 \times 1} = \frac{30}{2} = 15 \]

### Conclusion
Thus, the number of ways to distribute 7 identical balls into 3 boxes such that no box is empty is **15**.

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